The Peano axioms

This package formalizes [Tao06, chapter 2.1 - The Peano axioms] .

The report content.

𝗍𝗁𝖾 𝖯𝖾𝖺𝗇𝗈 𝖺𝗑𝗂𝗈𝗆𝗌

# 𝗧𝗵𝗲𝗼𝗿𝘆 𝗽𝗿𝗼𝗽𝗲𝗿𝘁𝗶𝗲𝘀
𝗖𝗼𝗻𝘀𝗶𝘀𝘁𝗲𝗻𝗰𝘆: undetermined
𝗦𝘁𝗮𝗯𝗶𝗹𝗶𝘇𝗲𝗱: False
𝗘𝘅𝘁𝗲𝗻𝗱𝗲𝗱 𝘁𝗵𝗲𝗼𝗿𝘆: N/A

# 𝗦𝗶𝗺𝗽𝗹𝗲-𝗼𝗯𝗷𝗲𝗰𝘁𝘀 𝗱𝗲𝗰𝗹𝗮𝗿𝗮𝘁𝗶𝗼𝗻𝘀
𝖫𝖾𝗍 ⌜𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟⌝, ⌜0⌝, ⌜1⌝, ⌜2⌝, ⌜3⌝, ⌜4⌝, ⌜5⌝, ⌜6⌝ 𝖻𝖾 𝑠𝑖𝑚𝑝𝑙𝑒-𝑜𝑏𝑗𝑒𝑐𝑡𝑠 𝗂𝗇 𝒰₄.

# 𝗥𝗲𝗹𝗮𝘁𝗶𝗼𝗻𝘀
𝖫𝖾𝗍 ⌜++⌝, ⌜𝐼𝑛𝑐⌝ 𝖻𝖾 𝑢𝑛𝑎𝑟𝑦-𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜⟹⌝, ⌜≠⌝, ⌜∧⌝, ⌜𝑖𝑠-𝑎⌝, ⌜=⌝ 𝖻𝖾 𝑏𝑖𝑛𝑎𝑟𝑦-𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠 𝗂𝗇 𝒰₄.

# 𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲𝘀
𝖳𝗁𝖾 𝖿𝗈𝗅𝗅𝗈𝗐𝗂𝗇𝗀 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾𝗌 𝖺𝗋𝖾 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗎𝗇𝖽𝖾𝗋 𝗍𝗁𝗂𝗌 𝗍𝗁𝖾𝗈𝗋𝗒:
𝖫𝖾𝗍 ⌜𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝓐, 𝐏 ⊢ 𝐏)⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₈, 𝐐₅ ⊢ (𝐏₈ ∧ 𝐐₅))⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲))⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄)⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁))⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑖𝑛𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑦-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛-2⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐏₉ = 𝐐₆), (𝐏₉ ≠ 𝐐₆) ⊢ 𝐼𝑛𝑐(𝒯₁))⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂)⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑝𝑟𝑜𝑜𝑓-𝑏𝑦-𝑟𝑒𝑓𝑢𝑡𝑎𝑡𝑖𝑜𝑛-2⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐇₁ 𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑡𝑒 (𝐱₇ = 𝐲₇)), 𝐼𝑛𝑐(𝐇₁) ⊢ (𝐱₇ ≠ 𝐲₇))⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₁, 𝐎₁ ⊢ 𝐐₁)⌝ 𝗂𝗇 𝒰₄.

# 𝗧𝗵𝗲𝗼𝗿𝘆 𝗲𝗹𝗮𝗯𝗼𝗿𝗮𝘁𝗶𝗼𝗻 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲
# 𝟮: 𝗧𝗵𝗲 𝗻𝗮𝘁𝘂𝗿𝗮𝗹 𝗻𝘂𝗺𝗯𝗲𝗿𝘀
## 𝟮.𝟭: 𝗧𝗵𝗲 𝗽𝗲𝗮𝗻𝗼 𝗮𝘅𝗶𝗼𝗺𝘀
### 𝗜𝗻𝗳𝗼𝗿𝗺𝗮𝗹 𝗱𝗲𝗳𝗶𝗻𝗶𝘁𝗶𝗼𝗻 𝗼𝗳 𝗻𝗮𝘁𝘂𝗿𝗮𝗹 𝗻𝘂𝗺𝗯𝗲𝗿
### 𝗔𝘅𝗶𝗼𝗺 𝟮.𝟭
𝗔𝘅𝗶𝗼𝗺 𝟮.𝟭 (𝒯₁.𝐴₁): 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₁ ⌜𝟢 𝘪𝘴 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳.⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝓐, 𝐏 ⊢ 𝐏)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁): (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟).
### 𝗔𝘅𝗶𝗼𝗺 𝟮.𝟮
𝗔𝘅𝗶𝗼𝗺 𝟮.𝟮 (𝒯₁.𝐴₂): 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₂ ⌜𝘐𝘧 𝘯 𝘪𝘴 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳, 𝘵𝘩𝘦𝘯 𝘯++ 𝘪𝘴 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳.⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂): ((𝐧₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₁)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)).
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₁, 𝐎₁ ⊢ 𝐐₁)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃): ((0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)).
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 𝟮.𝟮.𝟯 (𝒯₁.𝑃₄): ((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟).
𝗗𝗲𝗳𝗶𝗻𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝐷₁): 𝖫𝖾𝗍 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝒟₁ ⌜𝘞𝘦 𝘥𝘦𝘧𝘪𝘯𝘦 𝟣 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 𝟢++, 𝟤 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 (𝟢++)++, 𝟥 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 ((𝟢++)++)++,𝘦𝘵𝘤. (𝘐𝘯 𝘰𝘵𝘩𝘦𝘳 𝘸𝘰𝘳𝘥𝘴, 𝟣 := 𝟢++, 𝟤 := 𝟣++, 𝟥 := 𝟤++, 𝘦𝘵𝘤. 𝘐𝘯 𝘵𝘩𝘪𝘴 𝘵𝘦𝘹𝘵 𝘐 𝘶𝘴𝘦 "𝘹 := 𝘺" 𝘵𝘰 𝘥𝘦𝘯𝘰𝘵𝘦 𝘵𝘩𝘦 𝘴𝘵𝘢𝘵𝘦𝘮𝘦𝘯𝘵 𝘵𝘩𝘢𝘵 𝘹 𝘪𝘴 𝘥𝘦𝘧𝘪𝘯𝘦𝘥 𝘵𝘰 𝘦𝘲𝘶𝘢𝘭 𝘺.)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₅): (1 = (0)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₆): (2 = ((0)++)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇): (3 = (((0)++)++)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₈): (4 = ((((0)++)++)++)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₉): (((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₀): (((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₁): ((((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₂): ((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₃): (((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₄): (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟).
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₅): ((0)++ = 1).
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₆): (2 = (1)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₇): (((0)++)++ = 2).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₈): (3 = (2)++).
### 𝟯 𝗶𝘀 𝗮 𝗻𝗮𝘁𝘂𝗿𝗮𝗹 𝗻𝘂𝗺𝗯𝗲𝗿
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₉): ((((0)++)++)++ = 3).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₀): ((2)++ = 3).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 𝟮.𝟭.𝟰 (𝒯₁.𝑃₂₁): (3 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₂): (4 = ((((0)++)++)++)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₃): (((((0)++)++)++)++ = 4).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₄): ((3)++ = 4).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₅): (((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₆): (4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟).
### 𝗔𝘅𝗶𝗼𝗺 𝟮.𝟯
𝗔𝘅𝗶𝗼𝗺 𝟮.𝟯 (𝒯₁.𝐴₃): 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₃ ⌜𝟢 𝘪𝘴 𝘯𝘰𝘵 𝘵𝘩𝘦 𝘴𝘶𝘤𝘤𝘦𝘴𝘴𝘰𝘳 𝘰𝘧 𝘢𝘯𝘺 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳; 𝘪.𝘦., 𝘸𝘦 𝘩𝘢𝘷𝘦 𝘯++ ≠ 𝟢 𝘧𝘰𝘳 𝘦𝘷𝘦𝘳𝘺 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳 𝘯.⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₇): ((𝐧₂ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₂)++ ≠ 0)).
### 𝟰 𝗶𝘀 𝗻𝗼𝘁 𝗲𝗾𝘂𝗮𝗹 𝘁𝗼 𝟬.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₈): ((3 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((3)++ ≠ 0)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₉): ((3)++ ≠ 0).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 𝟮.𝟭.𝟲 (𝒯₁.𝑃₃₀): (4 ≠ 0).
### 𝗔𝘅𝗶𝗼𝗺 𝟮.𝟰
𝗔𝘅𝗶𝗼𝗺 𝟮.𝟰 (𝒯₁.𝐴₄): 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₄ ⌜𝘋𝘪𝘧𝘧𝘦𝘳𝘦𝘯𝘵 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳𝘴 𝘮𝘶𝘴𝘵 𝘩𝘢𝘷𝘦 𝘥𝘪𝘧𝘧𝘦𝘳𝘦𝘯𝘵 𝘴𝘶𝘤𝘤𝘦𝘴𝘴𝘰𝘳𝘴; 𝘪.𝘦., 𝘪𝘧 𝘯, 𝘮 𝘢𝘳𝘦 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳𝘴 𝘢𝘯𝘥 𝘯 ≠ 𝘮, 𝘵𝘩𝘦𝘯 𝘯++ ≠ 𝘮++. 𝘌𝘲𝘶𝘪𝘷𝘢𝘭𝘦𝘯𝘵𝘭𝘺, 𝘪𝘧 𝘯++ = 𝘮++, 𝘵𝘩𝘦𝘯 𝘸𝘦 𝘮𝘶𝘴𝘵 𝘩𝘢𝘷𝘦 𝘯 = 𝘮.⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₁): ((((𝐧₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (𝐧₃ ≠ 𝐦₁)) ⟹ ((𝐧₃)++ ≠ (𝐦₁)++)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₂): ((((𝐧₄ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₂ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ ((𝐧₄)++ = (𝐦₂)++)) ⟹ (𝐧₄ = 𝐦₂)).
### 𝟲 𝗶𝘀 𝗻𝗼𝘁 𝗲𝗾𝘂𝗮𝗹 𝘁𝗼 𝟮.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₃): ((((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (4 ≠ 0)) ⟹ ((4)++ ≠ (0)++)).
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₈, 𝐐₅ ⊢ (𝐏₈ ∧ 𝐐₅))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₄): ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₅): (((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (4 ≠ 0)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₆): ((4)++ ≠ (0)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₇): (5 = (((((0)++)++)++)++)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₈): ((((((0)++)++)++)++)++ = 5).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₉): ((4)++ = 5).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₀): (5 = (4)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₁): ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ ((5)++ ≠ (1)++)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₂): ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((4)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₃): ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₄): (5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₅): ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ ((5)++ ≠ (1)++)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₆): ((4)++ ≠ (0)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₇): (5 ≠ (0)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₈): (6 = ((((((0)++)++)++)++)++)++).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₉): (((((((0)++)++)++)++)++)++ = 6).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₅₀): (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₅₁): ((5)++ = 6).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₅₂): (6 = (5)++).
#### 𝗣𝗿𝗼𝗼𝗳 𝗯𝘆 𝗰𝗼𝗻𝘁𝗿𝗮𝗱𝗶𝗰𝘁𝗶𝗼𝗻
𝗛𝘆𝗽𝗼𝘁𝗵𝗲𝘀𝗶𝘀 (𝒯₁.𝐻₁): (6 = 2).
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑖𝑛𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑦-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛-2): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑖𝑛𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑦-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛-2 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐏₉ = 𝐐₆), (𝐏₉ ≠ 𝐐₆) ⊢ 𝐼𝑛𝑐(𝒯₁))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₆₆): 𝐼𝑛𝑐(ℋ₁).
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑝𝑟𝑜𝑜𝑓-𝑏𝑦-𝑟𝑒𝑓𝑢𝑡𝑎𝑡𝑖𝑜𝑛-2): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑝𝑟𝑜𝑜𝑓-𝑏𝑦-𝑟𝑒𝑓𝑢𝑡𝑎𝑡𝑖𝑜𝑛-2 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐇₁ 𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑡𝑒 (𝐱₇ = 𝐲₇)), 𝐼𝑛𝑐(𝐇₁) ⊢ (𝐱₇ ≠ 𝐲₇))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 𝟮.𝟭.𝟴 (𝒯₁.𝑃₆₇): (6 ≠ 2).
#### 𝗗𝗶𝗿𝗲𝗰𝘁 𝗽𝗿𝗼𝗼𝗳
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₆₈): ((1)++ = 2).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₆₉): (5 ≠ 1).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₀): ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ (6 ≠ (1)++)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₁): ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ (6 ≠ 2)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₂): ((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₃): (((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)).
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₄): (6 ≠ 2).
### 𝗔𝘅𝗶𝗼𝗺 𝟮.𝟱: 𝘁𝗵𝗲 𝗽𝗿𝗶𝗻𝗰𝗶𝗽𝗹𝗲 𝗼𝗳 𝗺𝗮𝘁𝗵𝗲𝗺𝗮𝘁𝗶𝗰𝗮𝗹 𝗶𝗻𝗱𝘂𝗰𝘁𝗶𝗼𝗻
𝗔𝘅𝗶𝗼𝗺 𝘀𝗰𝗵𝗲𝗺𝗮 (𝒯₁.𝐴₆) - Principle of mathematical induction: 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₆ ⌜𝘓𝘦𝘵 𝘗(𝘯) 𝘣𝘦 𝘢𝘯𝘺 𝘱𝘳𝘰𝘱𝘦𝘳𝘵𝘺 𝘱𝘦𝘳𝘵𝘢𝘪𝘯𝘪𝘯𝘨 𝘵𝘰 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳 𝘯. 𝘚𝘶𝘱𝘱𝘰𝘴𝘦 𝘵𝘩𝘢𝘵 𝘗(𝘖) 𝘪𝘴 𝘵𝘳𝘶𝘦, 𝘢𝘯𝘥 𝘴𝘶𝘱𝘱𝘰𝘴𝘦 𝘵𝘩𝘢𝘵 𝘸𝘩𝘦𝘯𝘦𝘷𝘦𝘳 𝘗(𝘯) 𝘪𝘴 𝘵𝘳𝘶𝘦, 𝘗(𝘯++) 𝘪𝘴 𝘢𝘭𝘴𝘰 𝘵𝘳𝘶𝘦. 𝘛𝘩𝘦𝘯 𝘗(𝘯) 𝘪𝘴 𝘵𝘳𝘶𝘦 𝘧𝘰𝘳 𝘦𝘷𝘦𝘳𝘺 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳 𝘯.⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₅): (((𝐧₅ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐏₁₀(0) ∧ (𝐏₁₀(𝐧₅) ⟹ 𝐏₁₀((𝐧₅)++)))) ⟹ ((𝐦₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ 𝐏₁₀(𝐦₃))).
### 𝗧𝗵𝗲 𝗻𝘂𝗺𝗯𝗲𝗿 𝘀𝘆𝘀𝘁𝗲𝗺 𝗻
### 𝗥𝗲𝗰𝘂𝗿𝘀𝗶𝘃𝗲 𝗱𝗲𝗳𝗶𝗻𝗶𝘁𝗶𝗼𝗻𝘀

The report content.

𝗍𝗁𝖾 𝖯𝖾𝖺𝗇𝗈 𝖺𝗑𝗂𝗈𝗆𝗌

# 𝗧𝗵𝗲𝗼𝗿𝘆 𝗽𝗿𝗼𝗽𝗲𝗿𝘁𝗶𝗲𝘀
𝗖𝗼𝗻𝘀𝗶𝘀𝘁𝗲𝗻𝗰𝘆: undetermined
𝗦𝘁𝗮𝗯𝗶𝗹𝗶𝘇𝗲𝗱: False
𝗘𝘅𝘁𝗲𝗻𝗱𝗲𝗱 𝘁𝗵𝗲𝗼𝗿𝘆: N/A

# 𝗦𝗶𝗺𝗽𝗹𝗲-𝗼𝗯𝗷𝗲𝗰𝘁𝘀 𝗱𝗲𝗰𝗹𝗮𝗿𝗮𝘁𝗶𝗼𝗻𝘀
𝖫𝖾𝗍 ⌜𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟⌝, ⌜0⌝, ⌜1⌝, ⌜2⌝, ⌜3⌝, ⌜4⌝, ⌜5⌝, ⌜6⌝ 𝖻𝖾 𝑠𝑖𝑚𝑝𝑙𝑒-𝑜𝑏𝑗𝑒𝑐𝑡𝑠 𝗂𝗇 𝒰₄.

# 𝗥𝗲𝗹𝗮𝘁𝗶𝗼𝗻𝘀
𝖫𝖾𝗍 ⌜++⌝, ⌜𝐼𝑛𝑐⌝ 𝖻𝖾 𝑢𝑛𝑎𝑟𝑦-𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜⟹⌝, ⌜≠⌝, ⌜∧⌝, ⌜𝑖𝑠-𝑎⌝, ⌜=⌝ 𝖻𝖾 𝑏𝑖𝑛𝑎𝑟𝑦-𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠 𝗂𝗇 𝒰₄.

# 𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲𝘀
𝖳𝗁𝖾 𝖿𝗈𝗅𝗅𝗈𝗐𝗂𝗇𝗀 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾𝗌 𝖺𝗋𝖾 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗎𝗇𝖽𝖾𝗋 𝗍𝗁𝗂𝗌 𝗍𝗁𝖾𝗈𝗋𝗒:
𝖫𝖾𝗍 ⌜𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝓐, 𝐏 ⊢ 𝐏)⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₈, 𝐐₅ ⊢ (𝐏₈ ∧ 𝐐₅))⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲))⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄)⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁))⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑖𝑛𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑦-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛-2⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐏₉ = 𝐐₆), (𝐏₉ ≠ 𝐐₆) ⊢ 𝐼𝑛𝑐(𝒯₁))⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂)⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑝𝑟𝑜𝑜𝑓-𝑏𝑦-𝑟𝑒𝑓𝑢𝑡𝑎𝑡𝑖𝑜𝑛-2⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐇₁ 𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑡𝑒 (𝐱₇ = 𝐲₇)), 𝐼𝑛𝑐(𝐇₁) ⊢ (𝐱₇ ≠ 𝐲₇))⌝ 𝗂𝗇 𝒰₄.
𝖫𝖾𝗍 ⌜𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛⌝ 𝖻𝖾 𝖺𝗇 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₁, 𝐎₁ ⊢ 𝐐₁)⌝ 𝗂𝗇 𝒰₄.

# 𝗧𝗵𝗲𝗼𝗿𝘆 𝗲𝗹𝗮𝗯𝗼𝗿𝗮𝘁𝗶𝗼𝗻 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲
# 𝟮: 𝗧𝗵𝗲 𝗻𝗮𝘁𝘂𝗿𝗮𝗹 𝗻𝘂𝗺𝗯𝗲𝗿𝘀
## 𝟮.𝟭: 𝗧𝗵𝗲 𝗽𝗲𝗮𝗻𝗼 𝗮𝘅𝗶𝗼𝗺𝘀
### 𝗜𝗻𝗳𝗼𝗿𝗺𝗮𝗹 𝗱𝗲𝗳𝗶𝗻𝗶𝘁𝗶𝗼𝗻 𝗼𝗳 𝗻𝗮𝘁𝘂𝗿𝗮𝗹 𝗻𝘂𝗺𝗯𝗲𝗿
### 𝗔𝘅𝗶𝗼𝗺 𝟮.𝟭
𝗔𝘅𝗶𝗼𝗺 𝟮.𝟭 (𝒯₁.𝐴₁): 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₁ ⌜𝟢 𝘪𝘴 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳.⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝓐, 𝐏 ⊢ 𝐏)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁): (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). 𝗣𝗿𝗼𝗼𝗳: ⌜𝟢 𝘪𝘴 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳.⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗮𝘅𝗶𝗼𝗺 𝟮.𝟭 (𝐴₁). (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) 𝗂𝗌 𝖺 𝗉𝗋𝗈𝗉𝗈𝗌𝗂𝗍𝗂𝗈𝗇𝖺𝗅 𝖿𝗈𝗋𝗆𝗎𝗅𝖺 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖾𝖽 𝖿𝗋𝗈𝗆 𝗍𝗁𝖺𝗍 𝖺𝗑𝗂𝗈𝗆. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓐, 𝐏 ⊢ 𝐏), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). ∎
### 𝗔𝘅𝗶𝗼𝗺 𝟮.𝟮
𝗔𝘅𝗶𝗼𝗺 𝟮.𝟮 (𝒯₁.𝐴₂): 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₂ ⌜𝘐𝘧 𝘯 𝘪𝘴 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳, 𝘵𝘩𝘦𝘯 𝘯++ 𝘪𝘴 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳.⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂): ((𝐧₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₁)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘐𝘧 𝘯 𝘪𝘴 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳, 𝘵𝘩𝘦𝘯 𝘯++ 𝘪𝘴 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳.⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗮𝘅𝗶𝗼𝗺 𝟮.𝟮 (𝐴₂). ((𝐧₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₁)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝗂𝗌 𝖺 𝗉𝗋𝗈𝗉𝗈𝗌𝗂𝗍𝗂𝗈𝗇𝖺𝗅 𝖿𝗈𝗋𝗆𝗎𝗅𝖺 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖾𝖽 𝖿𝗋𝗈𝗆 𝗍𝗁𝖺𝗍 𝖺𝗑𝗂𝗈𝗆. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓐, 𝐏 ⊢ 𝐏), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((𝐧₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₁)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). ∎
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₁, 𝐎₁ ⊢ 𝐐₁)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃): ((0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). 𝗣𝗿𝗼𝗼𝗳: ((𝐧₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₁)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂). 𝖫𝖾𝗍 𝐧₁ = 0. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₁, 𝐎₁ ⊢ 𝐐₁), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). ∎
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 𝟮.𝟮.𝟯 (𝒯₁.𝑃₄): ((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). 𝗣𝗿𝗼𝗼𝗳: ((0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃).(0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). ∎
𝗗𝗲𝗳𝗶𝗻𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝐷₁): 𝖫𝖾𝗍 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝒟₁ ⌜𝘞𝘦 𝘥𝘦𝘧𝘪𝘯𝘦 𝟣 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 𝟢++, 𝟤 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 (𝟢++)++, 𝟥 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 ((𝟢++)++)++,𝘦𝘵𝘤. (𝘐𝘯 𝘰𝘵𝘩𝘦𝘳 𝘸𝘰𝘳𝘥𝘴, 𝟣 := 𝟢++, 𝟤 := 𝟣++, 𝟥 := 𝟤++, 𝘦𝘵𝘤. 𝘐𝘯 𝘵𝘩𝘪𝘴 𝘵𝘦𝘹𝘵 𝘐 𝘶𝘴𝘦 "𝘹 := 𝘺" 𝘵𝘰 𝘥𝘦𝘯𝘰𝘵𝘦 𝘵𝘩𝘦 𝘴𝘵𝘢𝘵𝘦𝘮𝘦𝘯𝘵 𝘵𝘩𝘢𝘵 𝘹 𝘪𝘴 𝘥𝘦𝘧𝘪𝘯𝘦𝘥 𝘵𝘰 𝘦𝘲𝘶𝘢𝘭 𝘺.)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₅): (1 = (0)++). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘞𝘦 𝘥𝘦𝘧𝘪𝘯𝘦 𝟣 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 𝟢++, 𝟤 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 (𝟢++)++, 𝟥 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 ((𝟢++)++)++,𝘦𝘵𝘤. (𝘐𝘯 𝘰𝘵𝘩𝘦𝘳 𝘸𝘰𝘳𝘥𝘴, 𝟣 := 𝟢++, 𝟤 := 𝟣++, 𝟥 := 𝟤++, 𝘦𝘵𝘤. 𝘐𝘯 𝘵𝘩𝘪𝘴 𝘵𝘦𝘹𝘵 𝘐 𝘶𝘴𝘦 "𝘹 := 𝘺" 𝘵𝘰 𝘥𝘦𝘯𝘰𝘵𝘦 𝘵𝘩𝘦 𝘴𝘵𝘢𝘵𝘦𝘮𝘦𝘯𝘵 𝘵𝘩𝘢𝘵 𝘹 𝘪𝘴 𝘥𝘦𝘧𝘪𝘯𝘦𝘥 𝘵𝘰 𝘦𝘲𝘶𝘢𝘭 𝘺.)⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗱𝗲𝗳. (𝐷₁). 1 𝗂𝗌 𝖺𝗇 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖺𝗍𝗂𝗈𝗇 𝗈𝖿 𝗍𝗁𝖺𝗍 𝖽𝖾𝖿𝗂𝗇𝗂𝗍𝗂𝗈𝗇. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (1 = (0)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₆): (2 = ((0)++)++). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘞𝘦 𝘥𝘦𝘧𝘪𝘯𝘦 𝟣 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 𝟢++, 𝟤 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 (𝟢++)++, 𝟥 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 ((𝟢++)++)++,𝘦𝘵𝘤. (𝘐𝘯 𝘰𝘵𝘩𝘦𝘳 𝘸𝘰𝘳𝘥𝘴, 𝟣 := 𝟢++, 𝟤 := 𝟣++, 𝟥 := 𝟤++, 𝘦𝘵𝘤. 𝘐𝘯 𝘵𝘩𝘪𝘴 𝘵𝘦𝘹𝘵 𝘐 𝘶𝘴𝘦 "𝘹 := 𝘺" 𝘵𝘰 𝘥𝘦𝘯𝘰𝘵𝘦 𝘵𝘩𝘦 𝘴𝘵𝘢𝘵𝘦𝘮𝘦𝘯𝘵 𝘵𝘩𝘢𝘵 𝘹 𝘪𝘴 𝘥𝘦𝘧𝘪𝘯𝘦𝘥 𝘵𝘰 𝘦𝘲𝘶𝘢𝘭 𝘺.)⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗱𝗲𝗳. (𝐷₁). 2 𝗂𝗌 𝖺𝗇 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖺𝗍𝗂𝗈𝗇 𝗈𝖿 𝗍𝗁𝖺𝗍 𝖽𝖾𝖿𝗂𝗇𝗂𝗍𝗂𝗈𝗇. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (2 = ((0)++)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇): (3 = (((0)++)++)++). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘞𝘦 𝘥𝘦𝘧𝘪𝘯𝘦 𝟣 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 𝟢++, 𝟤 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 (𝟢++)++, 𝟥 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 ((𝟢++)++)++,𝘦𝘵𝘤. (𝘐𝘯 𝘰𝘵𝘩𝘦𝘳 𝘸𝘰𝘳𝘥𝘴, 𝟣 := 𝟢++, 𝟤 := 𝟣++, 𝟥 := 𝟤++, 𝘦𝘵𝘤. 𝘐𝘯 𝘵𝘩𝘪𝘴 𝘵𝘦𝘹𝘵 𝘐 𝘶𝘴𝘦 "𝘹 := 𝘺" 𝘵𝘰 𝘥𝘦𝘯𝘰𝘵𝘦 𝘵𝘩𝘦 𝘴𝘵𝘢𝘵𝘦𝘮𝘦𝘯𝘵 𝘵𝘩𝘢𝘵 𝘹 𝘪𝘴 𝘥𝘦𝘧𝘪𝘯𝘦𝘥 𝘵𝘰 𝘦𝘲𝘶𝘢𝘭 𝘺.)⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗱𝗲𝗳. (𝐷₁). 3 𝗂𝗌 𝖺𝗇 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖺𝗍𝗂𝗈𝗇 𝗈𝖿 𝗍𝗁𝖺𝗍 𝖽𝖾𝖿𝗂𝗇𝗂𝗍𝗂𝗈𝗇. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (3 = (((0)++)++)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₈): (4 = ((((0)++)++)++)++). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘞𝘦 𝘥𝘦𝘧𝘪𝘯𝘦 𝟣 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 𝟢++, 𝟤 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 (𝟢++)++, 𝟥 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 ((𝟢++)++)++,𝘦𝘵𝘤. (𝘐𝘯 𝘰𝘵𝘩𝘦𝘳 𝘸𝘰𝘳𝘥𝘴, 𝟣 := 𝟢++, 𝟤 := 𝟣++, 𝟥 := 𝟤++, 𝘦𝘵𝘤. 𝘐𝘯 𝘵𝘩𝘪𝘴 𝘵𝘦𝘹𝘵 𝘐 𝘶𝘴𝘦 "𝘹 := 𝘺" 𝘵𝘰 𝘥𝘦𝘯𝘰𝘵𝘦 𝘵𝘩𝘦 𝘴𝘵𝘢𝘵𝘦𝘮𝘦𝘯𝘵 𝘵𝘩𝘢𝘵 𝘹 𝘪𝘴 𝘥𝘦𝘧𝘪𝘯𝘦𝘥 𝘵𝘰 𝘦𝘲𝘶𝘢𝘭 𝘺.)⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗱𝗲𝗳. (𝐷₁). 4 𝗂𝗌 𝖺𝗇 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖺𝗍𝗂𝗈𝗇 𝗈𝖿 𝗍𝗁𝖺𝗍 𝖽𝖾𝖿𝗂𝗇𝗂𝗍𝗂𝗈𝗇. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (4 = ((((0)++)++)++)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₉): (((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). 𝗣𝗿𝗼𝗼𝗳: ((𝐧₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₁)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂). 𝖫𝖾𝗍 𝐧₁ = (0)++. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₁, 𝐎₁ ⊢ 𝐐₁), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₀): (((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). 𝗣𝗿𝗼𝗼𝗳: (((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₉).((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. 𝟮.𝟮.𝟯 (𝑃₄). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₁): ((((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). 𝗣𝗿𝗼𝗼𝗳: ((𝐧₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₁)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂). 𝖫𝖾𝗍 𝐧₁ = ((0)++)++. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₁, 𝐎₁ ⊢ 𝐐₁), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₂): ((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). 𝗣𝗿𝗼𝗼𝗳: ((((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₁).(((0)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₀). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₃): (((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). 𝗣𝗿𝗼𝗼𝗳: ((𝐧₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₁)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂). 𝖫𝖾𝗍 𝐧₁ = (((0)++)++)++. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₁, 𝐎₁ ⊢ 𝐐₁), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₄): (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). 𝗣𝗿𝗼𝗼𝗳: (((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₃).((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₂). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). ∎
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₅): ((0)++ = 1). 𝗣𝗿𝗼𝗼𝗳: (1 = (0)++) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₅).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((0)++ = 1). ∎
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₆): (2 = (1)++). 𝗣𝗿𝗼𝗼𝗳: (2 = ((0)++)++) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₆). ((0)++ = 1) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₅). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (2 = (1)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₇): (((0)++)++ = 2). 𝗣𝗿𝗼𝗼𝗳: (2 = ((0)++)++) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₆).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((0)++)++ = 2). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₈): (3 = (2)++). 𝗣𝗿𝗼𝗼𝗳: (3 = (((0)++)++)++) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₇). (((0)++)++ = 2) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₇). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (3 = (2)++). ∎
### 𝟯 𝗶𝘀 𝗮 𝗻𝗮𝘁𝘂𝗿𝗮𝗹 𝗻𝘂𝗺𝗯𝗲𝗿
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₁₉): ((((0)++)++)++ = 3). 𝗣𝗿𝗼𝗼𝗳: (3 = (((0)++)++)++) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₇).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((0)++)++)++ = 3). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₀): ((2)++ = 3). 𝗣𝗿𝗼𝗼𝗳: ((((0)++)++)++ = 3) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₉). (((0)++)++ = 2) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₇). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((2)++ = 3). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 𝟮.𝟭.𝟰 (𝒯₁.𝑃₂₁): (3 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). 𝗣𝗿𝗼𝗼𝗳: ((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₂). ((((0)++)++)++ = 3) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₉). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (3 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₂): (4 = ((((0)++)++)++)++). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘞𝘦 𝘥𝘦𝘧𝘪𝘯𝘦 𝟣 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 𝟢++, 𝟤 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 (𝟢++)++, 𝟥 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 ((𝟢++)++)++,𝘦𝘵𝘤. (𝘐𝘯 𝘰𝘵𝘩𝘦𝘳 𝘸𝘰𝘳𝘥𝘴, 𝟣 := 𝟢++, 𝟤 := 𝟣++, 𝟥 := 𝟤++, 𝘦𝘵𝘤. 𝘐𝘯 𝘵𝘩𝘪𝘴 𝘵𝘦𝘹𝘵 𝘐 𝘶𝘴𝘦 "𝘹 := 𝘺" 𝘵𝘰 𝘥𝘦𝘯𝘰𝘵𝘦 𝘵𝘩𝘦 𝘴𝘵𝘢𝘵𝘦𝘮𝘦𝘯𝘵 𝘵𝘩𝘢𝘵 𝘹 𝘪𝘴 𝘥𝘦𝘧𝘪𝘯𝘦𝘥 𝘵𝘰 𝘦𝘲𝘶𝘢𝘭 𝘺.)⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗱𝗲𝗳. (𝐷₁). 4 𝗂𝗌 𝖺𝗇 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖺𝗍𝗂𝗈𝗇 𝗈𝖿 𝗍𝗁𝖺𝗍 𝖽𝖾𝖿𝗂𝗇𝗂𝗍𝗂𝗈𝗇. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (4 = ((((0)++)++)++)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₃): (((((0)++)++)++)++ = 4). 𝗣𝗿𝗼𝗼𝗳: (4 = ((((0)++)++)++)++) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₈).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((((0)++)++)++)++ = 4). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₄): ((3)++ = 4). 𝗣𝗿𝗼𝗼𝗳: (((((0)++)++)++)++ = 4) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂₃). ((((0)++)++)++ = 3) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₉). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((3)++ = 4). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₅): (((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). 𝗣𝗿𝗼𝗼𝗳: (((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₃). ((3)++ = 4) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂₄). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((((0)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₆): (4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). 𝗣𝗿𝗼𝗼𝗳: (((((0)++)++)++)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₄). (((((0)++)++)++)++ = 4) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂₃). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). ∎
### 𝗔𝘅𝗶𝗼𝗺 𝟮.𝟯
𝗔𝘅𝗶𝗼𝗺 𝟮.𝟯 (𝒯₁.𝐴₃): 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₃ ⌜𝟢 𝘪𝘴 𝘯𝘰𝘵 𝘵𝘩𝘦 𝘴𝘶𝘤𝘤𝘦𝘴𝘴𝘰𝘳 𝘰𝘧 𝘢𝘯𝘺 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳; 𝘪.𝘦., 𝘸𝘦 𝘩𝘢𝘷𝘦 𝘯++ ≠ 𝟢 𝘧𝘰𝘳 𝘦𝘷𝘦𝘳𝘺 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳 𝘯.⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₇): ((𝐧₂ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₂)++ ≠ 0)). 𝗣𝗿𝗼𝗼𝗳: ⌜𝟢 𝘪𝘴 𝘯𝘰𝘵 𝘵𝘩𝘦 𝘴𝘶𝘤𝘤𝘦𝘴𝘴𝘰𝘳 𝘰𝘧 𝘢𝘯𝘺 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳; 𝘪.𝘦., 𝘸𝘦 𝘩𝘢𝘷𝘦 𝘯++ ≠ 𝟢 𝘧𝘰𝘳 𝘦𝘷𝘦𝘳𝘺 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳 𝘯.⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗮𝘅𝗶𝗼𝗺 𝟮.𝟯 (𝐴₃). ((𝐧₂ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₂)++ ≠ 0)) 𝗂𝗌 𝖺 𝗉𝗋𝗈𝗉𝗈𝗌𝗂𝗍𝗂𝗈𝗇𝖺𝗅 𝖿𝗈𝗋𝗆𝗎𝗅𝖺 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖾𝖽 𝖿𝗋𝗈𝗆 𝗍𝗁𝖺𝗍 𝖺𝗑𝗂𝗈𝗆. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓐, 𝐏 ⊢ 𝐏), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((𝐧₂ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₂)++ ≠ 0)). ∎
### 𝟰 𝗶𝘀 𝗻𝗼𝘁 𝗲𝗾𝘂𝗮𝗹 𝘁𝗼 𝟬.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₈): ((3 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((3)++ ≠ 0)). 𝗣𝗿𝗼𝗼𝗳: ((𝐧₂ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₂)++ ≠ 0)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂₇). 𝖫𝖾𝗍 𝐧₂ = 3. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₁, 𝐎₁ ⊢ 𝐐₁), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((3 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((3)++ ≠ 0)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂₉): ((3)++ ≠ 0). 𝗣𝗿𝗼𝗼𝗳: ((3 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((3)++ ≠ 0)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂₈).(3 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. 𝟮.𝟭.𝟰 (𝑃₂₁). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((3)++ ≠ 0). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 𝟮.𝟭.𝟲 (𝒯₁.𝑃₃₀): (4 ≠ 0). 𝗣𝗿𝗼𝗼𝗳: ((3)++ ≠ 0) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂₉). ((3)++ = 4) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂₄). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (4 ≠ 0). ∎
### 𝗔𝘅𝗶𝗼𝗺 𝟮.𝟰
𝗔𝘅𝗶𝗼𝗺 𝟮.𝟰 (𝒯₁.𝐴₄): 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₄ ⌜𝘋𝘪𝘧𝘧𝘦𝘳𝘦𝘯𝘵 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳𝘴 𝘮𝘶𝘴𝘵 𝘩𝘢𝘷𝘦 𝘥𝘪𝘧𝘧𝘦𝘳𝘦𝘯𝘵 𝘴𝘶𝘤𝘤𝘦𝘴𝘴𝘰𝘳𝘴; 𝘪.𝘦., 𝘪𝘧 𝘯, 𝘮 𝘢𝘳𝘦 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳𝘴 𝘢𝘯𝘥 𝘯 ≠ 𝘮, 𝘵𝘩𝘦𝘯 𝘯++ ≠ 𝘮++. 𝘌𝘲𝘶𝘪𝘷𝘢𝘭𝘦𝘯𝘵𝘭𝘺, 𝘪𝘧 𝘯++ = 𝘮++, 𝘵𝘩𝘦𝘯 𝘸𝘦 𝘮𝘶𝘴𝘵 𝘩𝘢𝘷𝘦 𝘯 = 𝘮.⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₁): ((((𝐧₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (𝐧₃ ≠ 𝐦₁)) ⟹ ((𝐧₃)++ ≠ (𝐦₁)++)). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘋𝘪𝘧𝘧𝘦𝘳𝘦𝘯𝘵 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳𝘴 𝘮𝘶𝘴𝘵 𝘩𝘢𝘷𝘦 𝘥𝘪𝘧𝘧𝘦𝘳𝘦𝘯𝘵 𝘴𝘶𝘤𝘤𝘦𝘴𝘴𝘰𝘳𝘴; 𝘪.𝘦., 𝘪𝘧 𝘯, 𝘮 𝘢𝘳𝘦 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳𝘴 𝘢𝘯𝘥 𝘯 ≠ 𝘮, 𝘵𝘩𝘦𝘯 𝘯++ ≠ 𝘮++. 𝘌𝘲𝘶𝘪𝘷𝘢𝘭𝘦𝘯𝘵𝘭𝘺, 𝘪𝘧 𝘯++ = 𝘮++, 𝘵𝘩𝘦𝘯 𝘸𝘦 𝘮𝘶𝘴𝘵 𝘩𝘢𝘷𝘦 𝘯 = 𝘮.⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗮𝘅𝗶𝗼𝗺 𝟮.𝟰 (𝐴₄). ((((𝐧₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (𝐧₃ ≠ 𝐦₁)) ⟹ ((𝐧₃)++ ≠ (𝐦₁)++)) 𝗂𝗌 𝖺 𝗉𝗋𝗈𝗉𝗈𝗌𝗂𝗍𝗂𝗈𝗇𝖺𝗅 𝖿𝗈𝗋𝗆𝗎𝗅𝖺 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖾𝖽 𝖿𝗋𝗈𝗆 𝗍𝗁𝖺𝗍 𝖺𝗑𝗂𝗈𝗆. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓐, 𝐏 ⊢ 𝐏), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((𝐧₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (𝐧₃ ≠ 𝐦₁)) ⟹ ((𝐧₃)++ ≠ (𝐦₁)++)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₂): ((((𝐧₄ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₂ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ ((𝐧₄)++ = (𝐦₂)++)) ⟹ (𝐧₄ = 𝐦₂)). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘋𝘪𝘧𝘧𝘦𝘳𝘦𝘯𝘵 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳𝘴 𝘮𝘶𝘴𝘵 𝘩𝘢𝘷𝘦 𝘥𝘪𝘧𝘧𝘦𝘳𝘦𝘯𝘵 𝘴𝘶𝘤𝘤𝘦𝘴𝘴𝘰𝘳𝘴; 𝘪.𝘦., 𝘪𝘧 𝘯, 𝘮 𝘢𝘳𝘦 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳𝘴 𝘢𝘯𝘥 𝘯 ≠ 𝘮, 𝘵𝘩𝘦𝘯 𝘯++ ≠ 𝘮++. 𝘌𝘲𝘶𝘪𝘷𝘢𝘭𝘦𝘯𝘵𝘭𝘺, 𝘪𝘧 𝘯++ = 𝘮++, 𝘵𝘩𝘦𝘯 𝘸𝘦 𝘮𝘶𝘴𝘵 𝘩𝘢𝘷𝘦 𝘯 = 𝘮.⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗮𝘅𝗶𝗼𝗺 𝟮.𝟰 (𝐴₄). ((((𝐧₄ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₂ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ ((𝐧₄)++ = (𝐦₂)++)) ⟹ (𝐧₄ = 𝐦₂)) 𝗂𝗌 𝖺 𝗉𝗋𝗈𝗉𝗈𝗌𝗂𝗍𝗂𝗈𝗇𝖺𝗅 𝖿𝗈𝗋𝗆𝗎𝗅𝖺 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖾𝖽 𝖿𝗋𝗈𝗆 𝗍𝗁𝖺𝗍 𝖺𝗑𝗂𝗈𝗆. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓐, 𝐏 ⊢ 𝐏), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((𝐧₄ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₂ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ ((𝐧₄)++ = (𝐦₂)++)) ⟹ (𝐧₄ = 𝐦₂)). ∎
### 𝟲 𝗶𝘀 𝗻𝗼𝘁 𝗲𝗾𝘂𝗮𝗹 𝘁𝗼 𝟮.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₃): ((((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (4 ≠ 0)) ⟹ ((4)++ ≠ (0)++)). 𝗣𝗿𝗼𝗼𝗳: ((((𝐧₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (𝐧₃ ≠ 𝐦₁)) ⟹ ((𝐧₃)++ ≠ (𝐦₁)++)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₁). 𝖫𝖾𝗍 𝐧₃ = 4, 𝐦₁ = 0. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₁, 𝐎₁ ⊢ 𝐐₁), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (4 ≠ 0)) ⟹ ((4)++ ≠ (0)++)). ∎
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜(𝐏₈, 𝐐₅ ⊢ (𝐏₈ ∧ 𝐐₅))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₄): ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). 𝗣𝗿𝗼𝗼𝗳: (4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟), 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗈𝗋𝗆 𝐏₈, 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂₆). (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟), 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗈𝗋𝗆 𝐐₅, 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₈, 𝐐₅ ⊢ (𝐏₈ ∧ 𝐐₅)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₅): (((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (4 ≠ 0)). 𝗣𝗿𝗼𝗼𝗳: ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)), 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗈𝗋𝗆 𝐏₈, 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₄). (4 ≠ 0), 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗈𝗋𝗆 𝐐₅, 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. 𝟮.𝟭.𝟲 (𝑃₃₀).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₈, 𝐐₅ ⊢ (𝐏₈ ∧ 𝐐₅)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (4 ≠ 0)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₆): ((4)++ ≠ (0)++). 𝗣𝗿𝗼𝗼𝗳: ((((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (4 ≠ 0)) ⟹ ((4)++ ≠ (0)++)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₃).(((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (4 ≠ 0)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₅). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((4)++ ≠ (0)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₇): (5 = (((((0)++)++)++)++)++). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘞𝘦 𝘥𝘦𝘧𝘪𝘯𝘦 𝟣 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 𝟢++, 𝟤 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 (𝟢++)++, 𝟥 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 ((𝟢++)++)++,𝘦𝘵𝘤. (𝘐𝘯 𝘰𝘵𝘩𝘦𝘳 𝘸𝘰𝘳𝘥𝘴, 𝟣 := 𝟢++, 𝟤 := 𝟣++, 𝟥 := 𝟤++, 𝘦𝘵𝘤. 𝘐𝘯 𝘵𝘩𝘪𝘴 𝘵𝘦𝘹𝘵 𝘐 𝘶𝘴𝘦 "𝘹 := 𝘺" 𝘵𝘰 𝘥𝘦𝘯𝘰𝘵𝘦 𝘵𝘩𝘦 𝘴𝘵𝘢𝘵𝘦𝘮𝘦𝘯𝘵 𝘵𝘩𝘢𝘵 𝘹 𝘪𝘴 𝘥𝘦𝘧𝘪𝘯𝘦𝘥 𝘵𝘰 𝘦𝘲𝘶𝘢𝘭 𝘺.)⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗱𝗲𝗳. (𝐷₁). 5 𝗂𝗌 𝖺𝗇 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖺𝗍𝗂𝗈𝗇 𝗈𝖿 𝗍𝗁𝖺𝗍 𝖽𝖾𝖿𝗂𝗇𝗂𝗍𝗂𝗈𝗇. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (5 = (((((0)++)++)++)++)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₈): ((((((0)++)++)++)++)++ = 5). 𝗣𝗿𝗼𝗼𝗳: (5 = (((((0)++)++)++)++)++) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₇).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((((0)++)++)++)++)++ = 5). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃₉): ((4)++ = 5). 𝗣𝗿𝗼𝗼𝗳: ((((((0)++)++)++)++)++ = 5) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₈). (((((0)++)++)++)++ = 4) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂₃). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((4)++ = 5). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₀): (5 = (4)++). 𝗣𝗿𝗼𝗼𝗳: ((4)++ = 5) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₉).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (5 = (4)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₁): ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ ((5)++ ≠ (1)++)). 𝗣𝗿𝗼𝗼𝗳: ((((𝐧₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (𝐧₃ ≠ 𝐦₁)) ⟹ ((𝐧₃)++ ≠ (𝐦₁)++)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₁). 𝖫𝖾𝗍 𝐧₃ = 5, 𝐦₁ = 1. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₁, 𝐎₁ ⊢ 𝐐₁), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ ((5)++ ≠ (1)++)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₂): ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((4)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). 𝗣𝗿𝗼𝗼𝗳: ((𝐧₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((𝐧₁)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂). 𝖫𝖾𝗍 𝐧₁ = 4. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₁, 𝐎₁ ⊢ 𝐐₁), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((4)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₃): ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). 𝗣𝗿𝗼𝗼𝗳: ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ ((4)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₄₂). ((4)++ = 5) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₉). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₄): (5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). 𝗣𝗿𝗼𝗼𝗳: ((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ (5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₄₃).(4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂₆). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₅): ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ ((5)++ ≠ (1)++)). 𝗣𝗿𝗼𝗼𝗳: ((((𝐧₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐦₁ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (𝐧₃ ≠ 𝐦₁)) ⟹ ((𝐧₃)++ ≠ (𝐦₁)++)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₁). 𝖫𝖾𝗍 𝐧₃ = 5, 𝐦₁ = 1. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₁, 𝐎₁ ⊢ 𝐐₁), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ ((5)++ ≠ (1)++)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₆): ((4)++ ≠ (0)++). 𝗣𝗿𝗼𝗼𝗳: ((((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (4 ≠ 0)) ⟹ ((4)++ ≠ (0)++)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₃).(((4 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (0 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (4 ≠ 0)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₅). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((4)++ ≠ (0)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₇): (5 ≠ (0)++). 𝗣𝗿𝗼𝗼𝗳: ((4)++ ≠ (0)++) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₄₆). ((4)++ = 5) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₉). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (5 ≠ (0)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₈): (6 = ((((((0)++)++)++)++)++)++). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘞𝘦 𝘥𝘦𝘧𝘪𝘯𝘦 𝟣 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 𝟢++, 𝟤 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 (𝟢++)++, 𝟥 𝘵𝘰 𝘣𝘦 𝘵𝘩𝘦 𝘯𝘶𝘮𝘣𝘦𝘳 ((𝟢++)++)++,𝘦𝘵𝘤. (𝘐𝘯 𝘰𝘵𝘩𝘦𝘳 𝘸𝘰𝘳𝘥𝘴, 𝟣 := 𝟢++, 𝟤 := 𝟣++, 𝟥 := 𝟤++, 𝘦𝘵𝘤. 𝘐𝘯 𝘵𝘩𝘪𝘴 𝘵𝘦𝘹𝘵 𝘐 𝘶𝘴𝘦 "𝘹 := 𝘺" 𝘵𝘰 𝘥𝘦𝘯𝘰𝘵𝘦 𝘵𝘩𝘦 𝘴𝘵𝘢𝘵𝘦𝘮𝘦𝘯𝘵 𝘵𝘩𝘢𝘵 𝘹 𝘪𝘴 𝘥𝘦𝘧𝘪𝘯𝘦𝘥 𝘵𝘰 𝘦𝘲𝘶𝘢𝘭 𝘺.)⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗱𝗲𝗳. (𝐷₁). 6 𝗂𝗌 𝖺𝗇 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖺𝗍𝗂𝗈𝗇 𝗈𝖿 𝗍𝗁𝖺𝗍 𝖽𝖾𝖿𝗂𝗇𝗂𝗍𝗂𝗈𝗇. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓓, 𝐱, 𝐲 ⊢ (𝐱 = 𝐲)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (6 = ((((((0)++)++)++)++)++)++). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₄₉): (((((((0)++)++)++)++)++)++ = 6). 𝗣𝗿𝗼𝗼𝗳: (6 = ((((((0)++)++)++)++)++)++) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₄₈).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((((((0)++)++)++)++)++)++ = 6). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₅₀): (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). 𝗣𝗿𝗼𝗼𝗳: ((0)++ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. 𝟮.𝟮.𝟯 (𝑃₄). ((0)++ = 1) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₅). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₅₁): ((5)++ = 6). 𝗣𝗿𝗼𝗼𝗳: (((((((0)++)++)++)++)++)++ = 6) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₄₉). ((((((0)++)++)++)++)++ = 5) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₃₈). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((5)++ = 6). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₅₂): (6 = (5)++). 𝗣𝗿𝗼𝗼𝗳: ((5)++ = 6) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₅₁).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦-𝑐𝑜𝑚𝑚𝑢𝑡𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐱₁ = 𝐲₁) ⊢ (𝐲₁ = 𝐱₁)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (6 = (5)++). ∎
#### 𝗣𝗿𝗼𝗼𝗳 𝗯𝘆 𝗰𝗼𝗻𝘁𝗿𝗮𝗱𝗶𝗰𝘁𝗶𝗼𝗻
𝗛𝘆𝗽𝗼𝘁𝗵𝗲𝘀𝗶𝘀 (𝒯₁.𝐻₁): (6 = 2). 𝖳𝗁𝗂𝗌 𝗁𝗒𝗉𝗈𝗍𝗁𝖾𝗌𝗂𝗌 𝗂𝗌 𝖾𝗅𝖺𝖻𝗈𝗋𝖺𝗍𝖾𝖽 𝗂𝗇 𝗍𝗁𝖾𝗈𝗋𝗒 ℋ₁.
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑖𝑛𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑦-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛-2): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑖𝑛𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑦-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛-2 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐏₉ = 𝐐₆), (𝐏₉ ≠ 𝐐₆) ⊢ 𝐼𝑛𝑐(𝒯₁))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₆₆): 𝐼𝑛𝑐(ℋ₁). 𝗣𝗿𝗼𝗼𝗳: 𝖫𝖾𝗍 (𝑷 = 𝑸) := (4 = 0) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₆₅). 𝖫𝖾𝗍 (𝑷 ≠ 𝑸) := (4 ≠ 0) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. 𝟮.𝟭.𝟲 (𝑃₃₀).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑖𝑛𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑦-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛-2 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₉ = 𝐐₆), (𝐏₉ ≠ 𝐐₆) ⊢ 𝐼𝑛𝑐(𝒯₁)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 𝐼𝑛𝑐(ℋ₁). ∎
𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑝𝑟𝑜𝑜𝑓-𝑏𝑦-𝑟𝑒𝑓𝑢𝑡𝑎𝑡𝑖𝑜𝑛-2): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑝𝑟𝑜𝑜𝑓-𝑏𝑦-𝑟𝑒𝑓𝑢𝑡𝑎𝑡𝑖𝑜𝑛-2 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐇₁ 𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑡𝑒 (𝐱₇ = 𝐲₇)), 𝐼𝑛𝑐(𝐇₁) ⊢ (𝐱₇ ≠ 𝐲₇))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 𝟮.𝟭.𝟴 (𝒯₁.𝑃₆₇): (6 ≠ 2). 𝗣𝗿𝗼𝗼𝗳: 𝖫𝖾𝗍 𝗵𝘆𝗽. (𝐻₁) 𝖻𝖾 𝗍𝗁𝖾 𝗁𝗒𝗉𝗈𝗍𝗁𝖾𝗌𝗂𝗌 (6 = 2). 𝐼𝑛𝑐(ℋ₁) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₆₆). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑝𝑟𝑜𝑜𝑓-𝑏𝑦-𝑟𝑒𝑓𝑢𝑡𝑎𝑡𝑖𝑜𝑛-2 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐇₁ 𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑡𝑒 (𝐱₇ = 𝐲₇)), 𝐼𝑛𝑐(𝐇₁) ⊢ (𝐱₇ ≠ 𝐲₇)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (6 ≠ 2). ∎
#### 𝗗𝗶𝗿𝗲𝗰𝘁 𝗽𝗿𝗼𝗼𝗳
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₆₈): ((1)++ = 2). 𝗣𝗿𝗼𝗼𝗳: (((0)++)++ = 2) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₇). ((0)++ = 1) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₅). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((1)++ = 2). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₆₉): (5 ≠ 1). 𝗣𝗿𝗼𝗼𝗳: (5 ≠ (0)++) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₄₇). ((0)++ = 1) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁₅). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (5 ≠ 1). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₀): ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ (6 ≠ (1)++)). 𝗣𝗿𝗼𝗼𝗳: ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ ((5)++ ≠ (1)++)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₄₅). ((5)++ = 6) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₅₁). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ (6 ≠ (1)++)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₁): ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ (6 ≠ 2)). 𝗣𝗿𝗼𝗼𝗳: ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ (6 ≠ (1)++)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₇₀). ((1)++ = 2) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₆₈). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑒𝑞𝑢𝑎𝑙-𝑡𝑒𝑟𝑚𝑠-𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₆, (𝐱₃ = 𝐲₃) ⊢ 𝐐₄), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ (6 ≠ 2)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₂): ((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). 𝗣𝗿𝗼𝗼𝗳: (5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟), 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗈𝗋𝗆 𝐏₈, 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₄₄). (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟), 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗈𝗋𝗆 𝐐₅, 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₅₀).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₈, 𝐐₅ ⊢ (𝐏₈ ∧ 𝐐₅)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 ((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₃): (((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)). 𝗣𝗿𝗼𝗼𝗳: ((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)), 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗈𝗋𝗆 𝐏₈, 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₇₂). (5 ≠ 1), 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗈𝗋𝗆 𝐐₅, 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₆₉).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑐𝑜𝑛𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛-𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝐏₈, 𝐐₅ ⊢ (𝐏₈ ∧ 𝐐₅)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)). ∎
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₄): (6 ≠ 2). 𝗣𝗿𝗼𝗼𝗳: ((((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) ⟹ (6 ≠ 2)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₇₁).(((5 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (1 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟)) ∧ (5 ≠ 1)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₇₃). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₃ ⟹ 𝐐₂), 𝐏₃ ⊢ 𝐐₂), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (6 ≠ 2). ∎
### 𝗔𝘅𝗶𝗼𝗺 𝟮.𝟱: 𝘁𝗵𝗲 𝗽𝗿𝗶𝗻𝗰𝗶𝗽𝗹𝗲 𝗼𝗳 𝗺𝗮𝘁𝗵𝗲𝗺𝗮𝘁𝗶𝗰𝗮𝗹 𝗶𝗻𝗱𝘂𝗰𝘁𝗶𝗼𝗻
𝗔𝘅𝗶𝗼𝗺 𝘀𝗰𝗵𝗲𝗺𝗮 (𝒯₁.𝐴₆) - Principle of mathematical induction: 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₆ ⌜𝘓𝘦𝘵 𝘗(𝘯) 𝘣𝘦 𝘢𝘯𝘺 𝘱𝘳𝘰𝘱𝘦𝘳𝘵𝘺 𝘱𝘦𝘳𝘵𝘢𝘪𝘯𝘪𝘯𝘨 𝘵𝘰 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳 𝘯. 𝘚𝘶𝘱𝘱𝘰𝘴𝘦 𝘵𝘩𝘢𝘵 𝘗(𝘖) 𝘪𝘴 𝘵𝘳𝘶𝘦, 𝘢𝘯𝘥 𝘴𝘶𝘱𝘱𝘰𝘴𝘦 𝘵𝘩𝘢𝘵 𝘸𝘩𝘦𝘯𝘦𝘷𝘦𝘳 𝘗(𝘯) 𝘪𝘴 𝘵𝘳𝘶𝘦, 𝘗(𝘯++) 𝘪𝘴 𝘢𝘭𝘴𝘰 𝘵𝘳𝘶𝘦. 𝘛𝘩𝘦𝘯 𝘗(𝘯) 𝘪𝘴 𝘵𝘳𝘶𝘦 𝘧𝘰𝘳 𝘦𝘷𝘦𝘳𝘺 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳 𝘯.⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 (𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽) 𝗂𝗇 𝒯₁.
𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₇₅): (((𝐧₅ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐏₁₀(0) ∧ (𝐏₁₀(𝐧₅) ⟹ 𝐏₁₀((𝐧₅)++)))) ⟹ ((𝐦₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ 𝐏₁₀(𝐦₃))). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘓𝘦𝘵 𝘗(𝘯) 𝘣𝘦 𝘢𝘯𝘺 𝘱𝘳𝘰𝘱𝘦𝘳𝘵𝘺 𝘱𝘦𝘳𝘵𝘢𝘪𝘯𝘪𝘯𝘨 𝘵𝘰 𝘢 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳 𝘯. 𝘚𝘶𝘱𝘱𝘰𝘴𝘦 𝘵𝘩𝘢𝘵 𝘗(𝘖) 𝘪𝘴 𝘵𝘳𝘶𝘦, 𝘢𝘯𝘥 𝘴𝘶𝘱𝘱𝘰𝘴𝘦 𝘵𝘩𝘢𝘵 𝘸𝘩𝘦𝘯𝘦𝘷𝘦𝘳 𝘗(𝘯) 𝘪𝘴 𝘵𝘳𝘶𝘦, 𝘗(𝘯++) 𝘪𝘴 𝘢𝘭𝘴𝘰 𝘵𝘳𝘶𝘦. 𝘛𝘩𝘦𝘯 𝘗(𝘯) 𝘪𝘴 𝘵𝘳𝘶𝘦 𝘧𝘰𝘳 𝘦𝘷𝘦𝘳𝘺 𝘯𝘢𝘵𝘶𝘳𝘢𝘭 𝘯𝘶𝘮𝘣𝘦𝘳 𝘯.⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗮𝘅𝗶𝗼𝗺 𝘀𝗰𝗵𝗲𝗺𝗮 (𝐴₆). (((𝐧₅ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐏₁₀(0) ∧ (𝐏₁₀(𝐧₅) ⟹ 𝐏₁₀((𝐧₅)++)))) ⟹ ((𝐦₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ 𝐏₁₀(𝐦₃))) 𝗂𝗌 𝖺 𝗉𝗋𝗈𝗉𝗈𝗌𝗂𝗍𝗂𝗈𝗇𝖺𝗅 𝖿𝗈𝗋𝗆𝗎𝗅𝖺 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖾𝖽 𝖿𝗋𝗈𝗆 𝗍𝗁𝖺𝗍 𝖺𝗑𝗂𝗈𝗆. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓐, 𝐏 ⊢ 𝐏), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (((𝐧₅ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ∧ (𝐏₁₀(0) ∧ (𝐏₁₀(𝐧₅) ⟹ 𝐏₁₀((𝐧₅)++)))) ⟹ ((𝐦₃ 𝑖𝑠-𝑎 𝑛𝑎𝑡𝑢𝑟𝑎𝑙-𝑛𝑢𝑚𝑏𝑒𝑟) ⟹ 𝐏₁₀(𝐦₃))). ∎
### 𝗧𝗵𝗲 𝗻𝘂𝗺𝗯𝗲𝗿 𝘀𝘆𝘀𝘁𝗲𝗺 𝗻
### 𝗥𝗲𝗰𝘂𝗿𝘀𝗶𝘃𝗲 𝗱𝗲𝗳𝗶𝗻𝗶𝘁𝗶𝗼𝗻𝘀

The report content.

𝗍𝗁𝖾 𝖯𝖾𝖺𝗇𝗈 𝖺𝗑𝗂𝗈𝗆𝗌

# Theory properties
Consistency: undetermined
Stabilized: False
Extended theory: N/A

# Simple-objects declarations
Let "natural-number", "0", "1", "2", "3", "4", "5", "6" be simple-objects in U4.

# Connectives
Let "++", "Inc" be unary-connectives in U4.
Let "==>", "neq", "and", "is-a", "=" be binary-connectives in U4.

# Inference rules
The following inference rules are considered valid under this theory:
Let "axiom-interpretation" be an inference-rule defined as "(A, P |- P)" in U4.
Let "conjunction-introduction" be an inference-rule defined as "(P8, Q5 |- (P8 and Q5))" in U4.
Let "definition-interpretation" be an inference-rule defined as "(D, x, y |- (x = y))" in U4.
Let "equal-terms-substitution" be an inference-rule defined as "(P6, (x3 = y3) |- Q4)" in U4.
Let "equality-commutativity" be an inference-rule defined as "((x1 = y1) |- (y1 = x1))" in U4.
Let "inconsistency-introduction-2" be an inference-rule defined as "((P9 = Q6), (P9 neq Q6) |- Inc(T1))" in U4.
Let "modus-ponens" be an inference-rule defined as "((P3 ==> Q2), P3 |- Q2)" in U4.
Let "proof-by-refutation-2" be an inference-rule defined as "((H1 formulate (x7 = y7)), Inc(H1) |- (x7 neq y7))" in U4.
Let "variable-substitution" be an inference-rule defined as "(P1, O1 |- Q1)" in U4.

# Theory elaboration sequence
# 2: The natural numbers
## 2.1: The peano axioms
### Informal definition of natural number
### Axiom 2.1
Axiom 2.1 (T1.A1): Let axiom A1 "0 is a natural number." be included (postulated) in T1.
Inference rule (axiom-interpretation): Let inference-rule axiom-interpretation defined as "(A, P |- P)" be included and considered valid in T1.
Proposition (T1.P1): (0 is-a natural-number).
### Axiom 2.2
Axiom 2.2 (T1.A2): Let axiom A2 "If n is a natural number, then n++ is a natural number." be included (postulated) in T1.
Proposition (T1.P2): ((n1 is-a natural-number) ==> ((n1)++ is-a natural-number)).
Inference rule (variable-substitution): Let inference-rule variable-substitution defined as "(P1, O1 |- Q1)" be included and considered valid in T1.
Proposition (T1.P3): ((0 is-a natural-number) ==> ((0)++ is-a natural-number)).
Inference rule (modus-ponens): Let inference-rule modus-ponens defined as "((P3 ==> Q2), P3 |- Q2)" be included and considered valid in T1.
Proposition 2.2.3 (T1.P4): ((0)++ is-a natural-number).
Definition (T1.D1): Let definition D1 "We define 1 to be the number 0++, 2 to be the number (0++)++, 3 to be the number ((0++)++)++,etc. (In other words, 1 := 0++, 2 := 1++, 3 := 2++, etc. In this text I use "x := y" to denote the statement that x is defined to equal y.)" be included (postulated) in T1.
Inference rule (definition-interpretation): Let inference-rule definition-interpretation defined as "(D, x, y |- (x = y))" be included and considered valid in T1.
Proposition (T1.P5): (1 = (0)++).
Proposition (T1.P6): (2 = ((0)++)++).
Proposition (T1.P7): (3 = (((0)++)++)++).
Proposition (T1.P8): (4 = ((((0)++)++)++)++).
Proposition (T1.P9): (((0)++ is-a natural-number) ==> (((0)++)++ is-a natural-number)).
Proposition (T1.P10): (((0)++)++ is-a natural-number).
Proposition (T1.P11): ((((0)++)++ is-a natural-number) ==> ((((0)++)++)++ is-a natural-number)).
Proposition (T1.P12): ((((0)++)++)++ is-a natural-number).
Proposition (T1.P13): (((((0)++)++)++ is-a natural-number) ==> (((((0)++)++)++)++ is-a natural-number)).
Proposition (T1.P14): (((((0)++)++)++)++ is-a natural-number).
Inference rule (equality-commutativity): Let inference-rule equality-commutativity defined as "((x1 = y1) |- (y1 = x1))" be included and considered valid in T1.
Proposition (T1.P15): ((0)++ = 1).
Inference rule (equal-terms-substitution): Let inference-rule equal-terms-substitution defined as "(P6, (x3 = y3) |- Q4)" be included and considered valid in T1.
Proposition (T1.P16): (2 = (1)++).
Proposition (T1.P17): (((0)++)++ = 2).
Proposition (T1.P18): (3 = (2)++).
### 3 is a natural number
Proposition (T1.P19): ((((0)++)++)++ = 3).
Proposition (T1.P20): ((2)++ = 3).
Proposition 2.1.4 (T1.P21): (3 is-a natural-number).
Proposition (T1.P22): (4 = ((((0)++)++)++)++).
Proposition (T1.P23): (((((0)++)++)++)++ = 4).
Proposition (T1.P24): ((3)++ = 4).
Proposition (T1.P25): (((((0)++)++)++ is-a natural-number) ==> (((((0)++)++)++)++ is-a natural-number)).
Proposition (T1.P26): (4 is-a natural-number).
### Axiom 2.3
Axiom 2.3 (T1.A3): Let axiom A3 "0 is not the successor of any natural number; i.e., we have n++  0 for every natural number n." be included (postulated) in T1.
Proposition (T1.P27): ((n2 is-a natural-number) ==> ((n2)++ neq 0)).
### 4 is not equal to 0.
Proposition (T1.P28): ((3 is-a natural-number) ==> ((3)++ neq 0)).
Proposition (T1.P29): ((3)++ neq 0).
Proposition 2.1.6 (T1.P30): (4 neq 0).
### Axiom 2.4
Axiom 2.4 (T1.A4): Let axiom A4 "Different natural numbers must have different successors; i.e., if n, m are natural numbers and n  m, then n++  m++. Equivalently, if n++ = m++, then we must have n = m." be included (postulated) in T1.
Proposition (T1.P31): ((((n3 is-a natural-number) and (m1 is-a natural-number)) and (n3 neq m1)) ==> ((n3)++ neq (m1)++)).
Proposition (T1.P32): ((((n4 is-a natural-number) and (m2 is-a natural-number)) and ((n4)++ = (m2)++)) ==> (n4 = m2)).
### 6 is not equal to 2.
Proposition (T1.P33): ((((4 is-a natural-number) and (0 is-a natural-number)) and (4 neq 0)) ==> ((4)++ neq (0)++)).
Inference rule (conjunction-introduction): Let inference-rule conjunction-introduction defined as "(P8, Q5 |- (P8 and Q5))" be included and considered valid in T1.
Proposition (T1.P34): ((4 is-a natural-number) and (0 is-a natural-number)).
Proposition (T1.P35): (((4 is-a natural-number) and (0 is-a natural-number)) and (4 neq 0)).
Proposition (T1.P36): ((4)++ neq (0)++).
Proposition (T1.P37): (5 = (((((0)++)++)++)++)++).
Proposition (T1.P38): ((((((0)++)++)++)++)++ = 5).
Proposition (T1.P39): ((4)++ = 5).
Proposition (T1.P40): (5 = (4)++).
Proposition (T1.P41): ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> ((5)++ neq (1)++)).
Proposition (T1.P42): ((4 is-a natural-number) ==> ((4)++ is-a natural-number)).
Proposition (T1.P43): ((4 is-a natural-number) ==> (5 is-a natural-number)).
Proposition (T1.P44): (5 is-a natural-number).
Proposition (T1.P45): ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> ((5)++ neq (1)++)).
Proposition (T1.P46): ((4)++ neq (0)++).
Proposition (T1.P47): (5 neq (0)++).
Proposition (T1.P48): (6 = ((((((0)++)++)++)++)++)++).
Proposition (T1.P49): (((((((0)++)++)++)++)++)++ = 6).
Proposition (T1.P50): (1 is-a natural-number).
Proposition (T1.P51): ((5)++ = 6).
Proposition (T1.P52): (6 = (5)++).
#### Proof by contradiction
Hypothesis (T1.H1): (6 = 2).
Inference rule (inconsistency-introduction-2): Let inference-rule inconsistency-introduction-2 defined as "((P9 = Q6), (P9 neq Q6) |- Inc(T1))" be included and considered valid in T1.
Proposition (T1.P66): Inc(H1).
Inference rule (proof-by-refutation-2): Let inference-rule proof-by-refutation-2 defined as "((H1 formulate (x7 = y7)), Inc(H1) |- (x7 neq y7))" be included and considered valid in T1.
Proposition 2.1.8 (T1.P67): (6 neq 2).
#### Direct proof
Proposition (T1.P68): ((1)++ = 2).
Proposition (T1.P69): (5 neq 1).
Proposition (T1.P70): ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> (6 neq (1)++)).
Proposition (T1.P71): ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> (6 neq 2)).
Proposition (T1.P72): ((5 is-a natural-number) and (1 is-a natural-number)).
Proposition (T1.P73): (((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)).
Proposition (T1.P74): (6 neq 2).
### Axiom 2.5: the principle of mathematical induction
Axiom schema (T1.A6) - Principle of mathematical induction: Let axiom A6 "Let P(n) be any property pertaining to a natural number n. Suppose that P(O) is true, and suppose that whenever P(n) is true, P(n++) is also true. Then P(n) is true for every natural number n." be included (postulated) in T1.
Proposition (T1.P75): (((n5 is-a natural-number) and (P10(0) and (P10(n5) ==> P10((n5)++)))) ==> ((m3 is-a natural-number) ==> P10(m3))).
### The number system n
### Recursive definitions

The report content.

𝗍𝗁𝖾 𝖯𝖾𝖺𝗇𝗈 𝖺𝗑𝗂𝗈𝗆𝗌

# Theory properties
Consistency: undetermined
Stabilized: False
Extended theory: N/A

# Simple-objects declarations
Let "natural-number", "0", "1", "2", "3", "4", "5", "6" be simple-objects in U4.

# Connectives
Let "++", "Inc" be unary-connectives in U4.
Let "==>", "neq", "and", "is-a", "=" be binary-connectives in U4.

# Inference rules
The following inference rules are considered valid under this theory:
Let "axiom-interpretation" be an inference-rule defined as "(A, P |- P)" in U4.
Let "conjunction-introduction" be an inference-rule defined as "(P8, Q5 |- (P8 and Q5))" in U4.
Let "definition-interpretation" be an inference-rule defined as "(D, x, y |- (x = y))" in U4.
Let "equal-terms-substitution" be an inference-rule defined as "(P6, (x3 = y3) |- Q4)" in U4.
Let "equality-commutativity" be an inference-rule defined as "((x1 = y1) |- (y1 = x1))" in U4.
Let "inconsistency-introduction-2" be an inference-rule defined as "((P9 = Q6), (P9 neq Q6) |- Inc(T1))" in U4.
Let "modus-ponens" be an inference-rule defined as "((P3 ==> Q2), P3 |- Q2)" in U4.
Let "proof-by-refutation-2" be an inference-rule defined as "((H1 formulate (x7 = y7)), Inc(H1) |- (x7 neq y7))" in U4.
Let "variable-substitution" be an inference-rule defined as "(P1, O1 |- Q1)" in U4.

# Theory elaboration sequence
# 2: The natural numbers
## 2.1: The peano axioms
### Informal definition of natural number
### Axiom 2.1
Axiom 2.1 (T1.A1): Let axiom A1 "0 is a natural number." be included (postulated) in T1.
Inference rule (axiom-interpretation): Let inference-rule axiom-interpretation defined as "(A, P |- P)" be included and considered valid in T1.
Proposition (T1.P1): (0 is-a natural-number). Proof: "0 is a natural number." is postulated by axiom 2.1 (A1). (0 is-a natural-number) is a propositional formula interpreted from that axiom. Therefore, by the axiom-interpretation inference rule: (A, P |- P), it follows that (0 is-a natural-number). QED
### Axiom 2.2
Axiom 2.2 (T1.A2): Let axiom A2 "If n is a natural number, then n++ is a natural number." be included (postulated) in T1.
Proposition (T1.P2): ((n1 is-a natural-number) ==> ((n1)++ is-a natural-number)). Proof: "If n is a natural number, then n++ is a natural number." is postulated by axiom 2.2 (A2). ((n1 is-a natural-number) ==> ((n1)++ is-a natural-number)) is a propositional formula interpreted from that axiom. Therefore, by the axiom-interpretation inference rule: (A, P |- P), it follows that ((n1 is-a natural-number) ==> ((n1)++ is-a natural-number)). QED
Inference rule (variable-substitution): Let inference-rule variable-substitution defined as "(P1, O1 |- Q1)" be included and considered valid in T1.
Proposition (T1.P3): ((0 is-a natural-number) ==> ((0)++ is-a natural-number)). Proof: ((n1 is-a natural-number) ==> ((n1)++ is-a natural-number)) follows from prop. (P2). Let n1 = 0. Therefore, by the variable-substitution inference rule: (P1, O1 |- Q1), it follows that ((0 is-a natural-number) ==> ((0)++ is-a natural-number)). QED
Inference rule (modus-ponens): Let inference-rule modus-ponens defined as "((P3 ==> Q2), P3 |- Q2)" be included and considered valid in T1.
Proposition 2.2.3 (T1.P4): ((0)++ is-a natural-number). Proof: ((0 is-a natural-number) ==> ((0)++ is-a natural-number)) follows from prop. (P3).(0 is-a natural-number) follows from prop. (P1). Therefore, by the modus-ponens inference rule: ((P3 ==> Q2), P3 |- Q2), it follows that ((0)++ is-a natural-number). QED
Definition (T1.D1): Let definition D1 "We define 1 to be the number 0++, 2 to be the number (0++)++, 3 to be the number ((0++)++)++,etc. (In other words, 1 := 0++, 2 := 1++, 3 := 2++, etc. In this text I use "x := y" to denote the statement that x is defined to equal y.)" be included (postulated) in T1.
Inference rule (definition-interpretation): Let inference-rule definition-interpretation defined as "(D, x, y |- (x = y))" be included and considered valid in T1.
Proposition (T1.P5): (1 = (0)++). Proof: "We define 1 to be the number 0++, 2 to be the number (0++)++, 3 to be the number ((0++)++)++,etc. (In other words, 1 := 0++, 2 := 1++, 3 := 2++, etc. In this text I use "x := y" to denote the statement that x is defined to equal y.)" is postulated by def. (D1). 1 is an interpretation of that definition. Therefore, by the definition-interpretation inference rule: (D, x, y |- (x = y)), it follows that (1 = (0)++). QED
Proposition (T1.P6): (2 = ((0)++)++). Proof: "We define 1 to be the number 0++, 2 to be the number (0++)++, 3 to be the number ((0++)++)++,etc. (In other words, 1 := 0++, 2 := 1++, 3 := 2++, etc. In this text I use "x := y" to denote the statement that x is defined to equal y.)" is postulated by def. (D1). 2 is an interpretation of that definition. Therefore, by the definition-interpretation inference rule: (D, x, y |- (x = y)), it follows that (2 = ((0)++)++). QED
Proposition (T1.P7): (3 = (((0)++)++)++). Proof: "We define 1 to be the number 0++, 2 to be the number (0++)++, 3 to be the number ((0++)++)++,etc. (In other words, 1 := 0++, 2 := 1++, 3 := 2++, etc. In this text I use "x := y" to denote the statement that x is defined to equal y.)" is postulated by def. (D1). 3 is an interpretation of that definition. Therefore, by the definition-interpretation inference rule: (D, x, y |- (x = y)), it follows that (3 = (((0)++)++)++). QED
Proposition (T1.P8): (4 = ((((0)++)++)++)++). Proof: "We define 1 to be the number 0++, 2 to be the number (0++)++, 3 to be the number ((0++)++)++,etc. (In other words, 1 := 0++, 2 := 1++, 3 := 2++, etc. In this text I use "x := y" to denote the statement that x is defined to equal y.)" is postulated by def. (D1). 4 is an interpretation of that definition. Therefore, by the definition-interpretation inference rule: (D, x, y |- (x = y)), it follows that (4 = ((((0)++)++)++)++). QED
Proposition (T1.P9): (((0)++ is-a natural-number) ==> (((0)++)++ is-a natural-number)). Proof: ((n1 is-a natural-number) ==> ((n1)++ is-a natural-number)) follows from prop. (P2). Let n1 = (0)++. Therefore, by the variable-substitution inference rule: (P1, O1 |- Q1), it follows that (((0)++ is-a natural-number) ==> (((0)++)++ is-a natural-number)). QED
Proposition (T1.P10): (((0)++)++ is-a natural-number). Proof: (((0)++ is-a natural-number) ==> (((0)++)++ is-a natural-number)) follows from prop. (P9).((0)++ is-a natural-number) follows from prop. 2.2.3 (P4). Therefore, by the modus-ponens inference rule: ((P3 ==> Q2), P3 |- Q2), it follows that (((0)++)++ is-a natural-number). QED
Proposition (T1.P11): ((((0)++)++ is-a natural-number) ==> ((((0)++)++)++ is-a natural-number)). Proof: ((n1 is-a natural-number) ==> ((n1)++ is-a natural-number)) follows from prop. (P2). Let n1 = ((0)++)++. Therefore, by the variable-substitution inference rule: (P1, O1 |- Q1), it follows that ((((0)++)++ is-a natural-number) ==> ((((0)++)++)++ is-a natural-number)). QED
Proposition (T1.P12): ((((0)++)++)++ is-a natural-number). Proof: ((((0)++)++ is-a natural-number) ==> ((((0)++)++)++ is-a natural-number)) follows from prop. (P11).(((0)++)++ is-a natural-number) follows from prop. (P10). Therefore, by the modus-ponens inference rule: ((P3 ==> Q2), P3 |- Q2), it follows that ((((0)++)++)++ is-a natural-number). QED
Proposition (T1.P13): (((((0)++)++)++ is-a natural-number) ==> (((((0)++)++)++)++ is-a natural-number)). Proof: ((n1 is-a natural-number) ==> ((n1)++ is-a natural-number)) follows from prop. (P2). Let n1 = (((0)++)++)++. Therefore, by the variable-substitution inference rule: (P1, O1 |- Q1), it follows that (((((0)++)++)++ is-a natural-number) ==> (((((0)++)++)++)++ is-a natural-number)). QED
Proposition (T1.P14): (((((0)++)++)++)++ is-a natural-number). Proof: (((((0)++)++)++ is-a natural-number) ==> (((((0)++)++)++)++ is-a natural-number)) follows from prop. (P13).((((0)++)++)++ is-a natural-number) follows from prop. (P12). Therefore, by the modus-ponens inference rule: ((P3 ==> Q2), P3 |- Q2), it follows that (((((0)++)++)++)++ is-a natural-number). QED
Inference rule (equality-commutativity): Let inference-rule equality-commutativity defined as "((x1 = y1) |- (y1 = x1))" be included and considered valid in T1.
Proposition (T1.P15): ((0)++ = 1). Proof: (1 = (0)++) follows from prop. (P5).  Therefore, by the equality-commutativity inference rule: ((x1 = y1) |- (y1 = x1)), it follows that ((0)++ = 1). QED
Inference rule (equal-terms-substitution): Let inference-rule equal-terms-substitution defined as "(P6, (x3 = y3) |- Q4)" be included and considered valid in T1.
Proposition (T1.P16): (2 = (1)++). Proof: (2 = ((0)++)++) follows from prop. (P6). ((0)++ = 1) follows from prop. (P15). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that (2 = (1)++). QED
Proposition (T1.P17): (((0)++)++ = 2). Proof: (2 = ((0)++)++) follows from prop. (P6).  Therefore, by the equality-commutativity inference rule: ((x1 = y1) |- (y1 = x1)), it follows that (((0)++)++ = 2). QED
Proposition (T1.P18): (3 = (2)++). Proof: (3 = (((0)++)++)++) follows from prop. (P7). (((0)++)++ = 2) follows from prop. (P17). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that (3 = (2)++). QED
### 3 is a natural number
Proposition (T1.P19): ((((0)++)++)++ = 3). Proof: (3 = (((0)++)++)++) follows from prop. (P7).  Therefore, by the equality-commutativity inference rule: ((x1 = y1) |- (y1 = x1)), it follows that ((((0)++)++)++ = 3). QED
Proposition (T1.P20): ((2)++ = 3). Proof: ((((0)++)++)++ = 3) follows from prop. (P19). (((0)++)++ = 2) follows from prop. (P17). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that ((2)++ = 3). QED
Proposition 2.1.4 (T1.P21): (3 is-a natural-number). Proof: ((((0)++)++)++ is-a natural-number) follows from prop. (P12). ((((0)++)++)++ = 3) follows from prop. (P19). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that (3 is-a natural-number). QED
Proposition (T1.P22): (4 = ((((0)++)++)++)++). Proof: "We define 1 to be the number 0++, 2 to be the number (0++)++, 3 to be the number ((0++)++)++,etc. (In other words, 1 := 0++, 2 := 1++, 3 := 2++, etc. In this text I use "x := y" to denote the statement that x is defined to equal y.)" is postulated by def. (D1). 4 is an interpretation of that definition. Therefore, by the definition-interpretation inference rule: (D, x, y |- (x = y)), it follows that (4 = ((((0)++)++)++)++). QED
Proposition (T1.P23): (((((0)++)++)++)++ = 4). Proof: (4 = ((((0)++)++)++)++) follows from prop. (P8).  Therefore, by the equality-commutativity inference rule: ((x1 = y1) |- (y1 = x1)), it follows that (((((0)++)++)++)++ = 4). QED
Proposition (T1.P24): ((3)++ = 4). Proof: (((((0)++)++)++)++ = 4) follows from prop. (P23). ((((0)++)++)++ = 3) follows from prop. (P19). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that ((3)++ = 4). QED
Proposition (T1.P25): (((((0)++)++)++ is-a natural-number) ==> (((((0)++)++)++)++ is-a natural-number)). Proof: (((((0)++)++)++ is-a natural-number) ==> (((((0)++)++)++)++ is-a natural-number)) follows from prop. (P13). ((3)++ = 4) follows from prop. (P24). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that (((((0)++)++)++ is-a natural-number) ==> (((((0)++)++)++)++ is-a natural-number)). QED
Proposition (T1.P26): (4 is-a natural-number). Proof: (((((0)++)++)++)++ is-a natural-number) follows from prop. (P14). (((((0)++)++)++)++ = 4) follows from prop. (P23). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that (4 is-a natural-number). QED
### Axiom 2.3
Axiom 2.3 (T1.A3): Let axiom A3 "0 is not the successor of any natural number; i.e., we have n++  0 for every natural number n." be included (postulated) in T1.
Proposition (T1.P27): ((n2 is-a natural-number) ==> ((n2)++ neq 0)). Proof: "0 is not the successor of any natural number; i.e., we have n++  0 for every natural number n." is postulated by axiom 2.3 (A3). ((n2 is-a natural-number) ==> ((n2)++ neq 0)) is a propositional formula interpreted from that axiom. Therefore, by the axiom-interpretation inference rule: (A, P |- P), it follows that ((n2 is-a natural-number) ==> ((n2)++ neq 0)). QED
### 4 is not equal to 0.
Proposition (T1.P28): ((3 is-a natural-number) ==> ((3)++ neq 0)). Proof: ((n2 is-a natural-number) ==> ((n2)++ neq 0)) follows from prop. (P27). Let n2 = 3. Therefore, by the variable-substitution inference rule: (P1, O1 |- Q1), it follows that ((3 is-a natural-number) ==> ((3)++ neq 0)). QED
Proposition (T1.P29): ((3)++ neq 0). Proof: ((3 is-a natural-number) ==> ((3)++ neq 0)) follows from prop. (P28).(3 is-a natural-number) follows from prop. 2.1.4 (P21). Therefore, by the modus-ponens inference rule: ((P3 ==> Q2), P3 |- Q2), it follows that ((3)++ neq 0). QED
Proposition 2.1.6 (T1.P30): (4 neq 0). Proof: ((3)++ neq 0) follows from prop. (P29). ((3)++ = 4) follows from prop. (P24). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that (4 neq 0). QED
### Axiom 2.4
Axiom 2.4 (T1.A4): Let axiom A4 "Different natural numbers must have different successors; i.e., if n, m are natural numbers and n  m, then n++  m++. Equivalently, if n++ = m++, then we must have n = m." be included (postulated) in T1.
Proposition (T1.P31): ((((n3 is-a natural-number) and (m1 is-a natural-number)) and (n3 neq m1)) ==> ((n3)++ neq (m1)++)). Proof: "Different natural numbers must have different successors; i.e., if n, m are natural numbers and n  m, then n++  m++. Equivalently, if n++ = m++, then we must have n = m." is postulated by axiom 2.4 (A4). ((((n3 is-a natural-number) and (m1 is-a natural-number)) and (n3 neq m1)) ==> ((n3)++ neq (m1)++)) is a propositional formula interpreted from that axiom. Therefore, by the axiom-interpretation inference rule: (A, P |- P), it follows that ((((n3 is-a natural-number) and (m1 is-a natural-number)) and (n3 neq m1)) ==> ((n3)++ neq (m1)++)). QED
Proposition (T1.P32): ((((n4 is-a natural-number) and (m2 is-a natural-number)) and ((n4)++ = (m2)++)) ==> (n4 = m2)). Proof: "Different natural numbers must have different successors; i.e., if n, m are natural numbers and n  m, then n++  m++. Equivalently, if n++ = m++, then we must have n = m." is postulated by axiom 2.4 (A4). ((((n4 is-a natural-number) and (m2 is-a natural-number)) and ((n4)++ = (m2)++)) ==> (n4 = m2)) is a propositional formula interpreted from that axiom. Therefore, by the axiom-interpretation inference rule: (A, P |- P), it follows that ((((n4 is-a natural-number) and (m2 is-a natural-number)) and ((n4)++ = (m2)++)) ==> (n4 = m2)). QED
### 6 is not equal to 2.
Proposition (T1.P33): ((((4 is-a natural-number) and (0 is-a natural-number)) and (4 neq 0)) ==> ((4)++ neq (0)++)). Proof: ((((n3 is-a natural-number) and (m1 is-a natural-number)) and (n3 neq m1)) ==> ((n3)++ neq (m1)++)) follows from prop. (P31). Let n3 = 4, m1 = 0. Therefore, by the variable-substitution inference rule: (P1, O1 |- Q1), it follows that ((((4 is-a natural-number) and (0 is-a natural-number)) and (4 neq 0)) ==> ((4)++ neq (0)++)). QED
Inference rule (conjunction-introduction): Let inference-rule conjunction-introduction defined as "(P8, Q5 |- (P8 and Q5))" be included and considered valid in T1.
Proposition (T1.P34): ((4 is-a natural-number) and (0 is-a natural-number)). Proof: (4 is-a natural-number), of the form P8, follows from prop. (P26). (0 is-a natural-number), of the form Q5, follows from prop. (P1).  Therefore, by the conjunction-introduction inference rule: (P8, Q5 |- (P8 and Q5)), it follows that ((4 is-a natural-number) and (0 is-a natural-number)). QED
Proposition (T1.P35): (((4 is-a natural-number) and (0 is-a natural-number)) and (4 neq 0)). Proof: ((4 is-a natural-number) and (0 is-a natural-number)), of the form P8, follows from prop. (P34). (4 neq 0), of the form Q5, follows from prop. 2.1.6 (P30).  Therefore, by the conjunction-introduction inference rule: (P8, Q5 |- (P8 and Q5)), it follows that (((4 is-a natural-number) and (0 is-a natural-number)) and (4 neq 0)). QED
Proposition (T1.P36): ((4)++ neq (0)++). Proof: ((((4 is-a natural-number) and (0 is-a natural-number)) and (4 neq 0)) ==> ((4)++ neq (0)++)) follows from prop. (P33).(((4 is-a natural-number) and (0 is-a natural-number)) and (4 neq 0)) follows from prop. (P35). Therefore, by the modus-ponens inference rule: ((P3 ==> Q2), P3 |- Q2), it follows that ((4)++ neq (0)++). QED
Proposition (T1.P37): (5 = (((((0)++)++)++)++)++). Proof: "We define 1 to be the number 0++, 2 to be the number (0++)++, 3 to be the number ((0++)++)++,etc. (In other words, 1 := 0++, 2 := 1++, 3 := 2++, etc. In this text I use "x := y" to denote the statement that x is defined to equal y.)" is postulated by def. (D1). 5 is an interpretation of that definition. Therefore, by the definition-interpretation inference rule: (D, x, y |- (x = y)), it follows that (5 = (((((0)++)++)++)++)++). QED
Proposition (T1.P38): ((((((0)++)++)++)++)++ = 5). Proof: (5 = (((((0)++)++)++)++)++) follows from prop. (P37).  Therefore, by the equality-commutativity inference rule: ((x1 = y1) |- (y1 = x1)), it follows that ((((((0)++)++)++)++)++ = 5). QED
Proposition (T1.P39): ((4)++ = 5). Proof: ((((((0)++)++)++)++)++ = 5) follows from prop. (P38). (((((0)++)++)++)++ = 4) follows from prop. (P23). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that ((4)++ = 5). QED
Proposition (T1.P40): (5 = (4)++). Proof: ((4)++ = 5) follows from prop. (P39).  Therefore, by the equality-commutativity inference rule: ((x1 = y1) |- (y1 = x1)), it follows that (5 = (4)++). QED
Proposition (T1.P41): ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> ((5)++ neq (1)++)). Proof: ((((n3 is-a natural-number) and (m1 is-a natural-number)) and (n3 neq m1)) ==> ((n3)++ neq (m1)++)) follows from prop. (P31). Let n3 = 5, m1 = 1. Therefore, by the variable-substitution inference rule: (P1, O1 |- Q1), it follows that ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> ((5)++ neq (1)++)). QED
Proposition (T1.P42): ((4 is-a natural-number) ==> ((4)++ is-a natural-number)). Proof: ((n1 is-a natural-number) ==> ((n1)++ is-a natural-number)) follows from prop. (P2). Let n1 = 4. Therefore, by the variable-substitution inference rule: (P1, O1 |- Q1), it follows that ((4 is-a natural-number) ==> ((4)++ is-a natural-number)). QED
Proposition (T1.P43): ((4 is-a natural-number) ==> (5 is-a natural-number)). Proof: ((4 is-a natural-number) ==> ((4)++ is-a natural-number)) follows from prop. (P42). ((4)++ = 5) follows from prop. (P39). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that ((4 is-a natural-number) ==> (5 is-a natural-number)). QED
Proposition (T1.P44): (5 is-a natural-number). Proof: ((4 is-a natural-number) ==> (5 is-a natural-number)) follows from prop. (P43).(4 is-a natural-number) follows from prop. (P26). Therefore, by the modus-ponens inference rule: ((P3 ==> Q2), P3 |- Q2), it follows that (5 is-a natural-number). QED
Proposition (T1.P45): ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> ((5)++ neq (1)++)). Proof: ((((n3 is-a natural-number) and (m1 is-a natural-number)) and (n3 neq m1)) ==> ((n3)++ neq (m1)++)) follows from prop. (P31). Let n3 = 5, m1 = 1. Therefore, by the variable-substitution inference rule: (P1, O1 |- Q1), it follows that ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> ((5)++ neq (1)++)). QED
Proposition (T1.P46): ((4)++ neq (0)++). Proof: ((((4 is-a natural-number) and (0 is-a natural-number)) and (4 neq 0)) ==> ((4)++ neq (0)++)) follows from prop. (P33).(((4 is-a natural-number) and (0 is-a natural-number)) and (4 neq 0)) follows from prop. (P35). Therefore, by the modus-ponens inference rule: ((P3 ==> Q2), P3 |- Q2), it follows that ((4)++ neq (0)++). QED
Proposition (T1.P47): (5 neq (0)++). Proof: ((4)++ neq (0)++) follows from prop. (P46). ((4)++ = 5) follows from prop. (P39). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that (5 neq (0)++). QED
Proposition (T1.P48): (6 = ((((((0)++)++)++)++)++)++). Proof: "We define 1 to be the number 0++, 2 to be the number (0++)++, 3 to be the number ((0++)++)++,etc. (In other words, 1 := 0++, 2 := 1++, 3 := 2++, etc. In this text I use "x := y" to denote the statement that x is defined to equal y.)" is postulated by def. (D1). 6 is an interpretation of that definition. Therefore, by the definition-interpretation inference rule: (D, x, y |- (x = y)), it follows that (6 = ((((((0)++)++)++)++)++)++). QED
Proposition (T1.P49): (((((((0)++)++)++)++)++)++ = 6). Proof: (6 = ((((((0)++)++)++)++)++)++) follows from prop. (P48).  Therefore, by the equality-commutativity inference rule: ((x1 = y1) |- (y1 = x1)), it follows that (((((((0)++)++)++)++)++)++ = 6). QED
Proposition (T1.P50): (1 is-a natural-number). Proof: ((0)++ is-a natural-number) follows from prop. 2.2.3 (P4). ((0)++ = 1) follows from prop. (P15). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that (1 is-a natural-number). QED
Proposition (T1.P51): ((5)++ = 6). Proof: (((((((0)++)++)++)++)++)++ = 6) follows from prop. (P49). ((((((0)++)++)++)++)++ = 5) follows from prop. (P38). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that ((5)++ = 6). QED
Proposition (T1.P52): (6 = (5)++). Proof: ((5)++ = 6) follows from prop. (P51).  Therefore, by the equality-commutativity inference rule: ((x1 = y1) |- (y1 = x1)), it follows that (6 = (5)++). QED
#### Proof by contradiction
Hypothesis (T1.H1): (6 = 2). This hypothesis is elaborated in theory H1.
Inference rule (inconsistency-introduction-2): Let inference-rule inconsistency-introduction-2 defined as "((P9 = Q6), (P9 neq Q6) |- Inc(T1))" be included and considered valid in T1.
Proposition (T1.P66): Inc(H1). Proof: Let (P = Q) := (4 = 0) follows from prop. (P65). Let (P neq Q)) := (4 neq 0) follows from prop. 2.1.6 (P30).  Therefore, by the inconsistency-introduction-2 inference rule: ((P9 = Q6), (P9 neq Q6) |- Inc(T1)), it follows that Inc(H1). QED
Inference rule (proof-by-refutation-2): Let inference-rule proof-by-refutation-2 defined as "((H1 formulate (x7 = y7)), Inc(H1) |- (x7 neq y7))" be included and considered valid in T1.
Proposition 2.1.8 (T1.P67): (6 neq 2). Proof: Let hyp. (H1) be the hypothesis (6 = 2). Inc(H1) follows from prop. (P66). Therefore, by the proof-by-refutation-2 inference rule: ((H1 formulate (x7 = y7)), Inc(H1) |- (x7 neq y7)), it follows that (6 neq 2). QED
#### Direct proof
Proposition (T1.P68): ((1)++ = 2). Proof: (((0)++)++ = 2) follows from prop. (P17). ((0)++ = 1) follows from prop. (P15). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that ((1)++ = 2). QED
Proposition (T1.P69): (5 neq 1). Proof: (5 neq (0)++) follows from prop. (P47). ((0)++ = 1) follows from prop. (P15). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that (5 neq 1). QED
Proposition (T1.P70): ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> (6 neq (1)++)). Proof: ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> ((5)++ neq (1)++)) follows from prop. (P45). ((5)++ = 6) follows from prop. (P51). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> (6 neq (1)++)). QED
Proposition (T1.P71): ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> (6 neq 2)). Proof: ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> (6 neq (1)++)) follows from prop. (P70). ((1)++ = 2) follows from prop. (P68). Therefore, by the equal-terms-substitution inference rule: (P6, (x3 = y3) |- Q4), it follows that ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> (6 neq 2)). QED
Proposition (T1.P72): ((5 is-a natural-number) and (1 is-a natural-number)). Proof: (5 is-a natural-number), of the form P8, follows from prop. (P44). (1 is-a natural-number), of the form Q5, follows from prop. (P50).  Therefore, by the conjunction-introduction inference rule: (P8, Q5 |- (P8 and Q5)), it follows that ((5 is-a natural-number) and (1 is-a natural-number)). QED
Proposition (T1.P73): (((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)). Proof: ((5 is-a natural-number) and (1 is-a natural-number)), of the form P8, follows from prop. (P72). (5 neq 1), of the form Q5, follows from prop. (P69).  Therefore, by the conjunction-introduction inference rule: (P8, Q5 |- (P8 and Q5)), it follows that (((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)). QED
Proposition (T1.P74): (6 neq 2). Proof: ((((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) ==> (6 neq 2)) follows from prop. (P71).(((5 is-a natural-number) and (1 is-a natural-number)) and (5 neq 1)) follows from prop. (P73). Therefore, by the modus-ponens inference rule: ((P3 ==> Q2), P3 |- Q2), it follows that (6 neq 2). QED
### Axiom 2.5: the principle of mathematical induction
Axiom schema (T1.A6) - Principle of mathematical induction: Let axiom A6 "Let P(n) be any property pertaining to a natural number n. Suppose that P(O) is true, and suppose that whenever P(n) is true, P(n++) is also true. Then P(n) is true for every natural number n." be included (postulated) in T1.
Proposition (T1.P75): (((n5 is-a natural-number) and (P10(0) and (P10(n5) ==> P10((n5)++)))) ==> ((m3 is-a natural-number) ==> P10(m3))). Proof: "Let P(n) be any property pertaining to a natural number n. Suppose that P(O) is true, and suppose that whenever P(n) is true, P(n++) is also true. Then P(n) is true for every natural number n." is postulated by axiom schema (A6). (((n5 is-a natural-number) and (P10(0) and (P10(n5) ==> P10((n5)++)))) ==> ((m3 is-a natural-number) ==> P10(m3))) is a propositional formula interpreted from that axiom. Therefore, by the axiom-interpretation inference rule: (A, P |- P), it follows that (((n5 is-a natural-number) and (P10(0) and (P10(n5) ==> P10((n5)++)))) ==> ((m3 is-a natural-number) ==> P10(m3))). QED
### The number system n
### Recursive definitions