biconditional-elimination-2 (python sample)

Tags: biconditional-elimination-2 python sample

Usage

Call the infer_statement method from the inference-rule inclusion class listed in the i (unabridged inference_rules ) property of the theory-derivation :

u = pu.create_universe()
t = u.t()
...
# some theory derivation code
...
t.i.biconditional_elimination_2.infer_statement(...)

Sample code

Code output

𝖫𝖾𝗍 ⌜𝒰₈⌝ 𝖻𝖾 𝖺 𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒-𝑜𝑓-𝑑𝑖𝑠𝑐𝑜𝑢𝑟𝑠𝑒.

𝖫𝖾𝗍 ⌜𝒯₁⌝ 𝖻𝖾 𝖺 𝑡ℎ𝑒𝑜𝑟𝑦-𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑜𝑛 𝗂𝗇 𝒰₈.

𝗔𝘅𝗶𝗼𝗺 (𝒯₁.𝐴₁): 𝖫𝖾𝗍 𝑎𝑥𝑖𝑜𝑚 𝒜₁ ⌜𝘋𝘶𝘮𝘮𝘺 𝘢𝘹𝘪𝘰𝘮 𝘧𝘰𝘳 𝘥𝘦𝘮𝘰𝘯𝘴𝘵𝘳𝘢𝘵𝘪𝘰𝘯 𝘱𝘶𝘳𝘱𝘰𝘴𝘦𝘴⌝ 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): (𝑟₁(𝑜₁, 𝑜₂) ⟺ 𝑟₂(𝑜₃)). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘋𝘶𝘮𝘮𝘺 𝘢𝘹𝘪𝘰𝘮 𝘧𝘰𝘳 𝘥𝘦𝘮𝘰𝘯𝘴𝘵𝘳𝘢𝘵𝘪𝘰𝘯 𝘱𝘶𝘳𝘱𝘰𝘴𝘦𝘴⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗮𝘅𝗶𝗼𝗺 (𝐴₁). (𝑟₁(𝑜₁, 𝑜₂) ⟺ 𝑟₂(𝑜₃)) 𝗂𝗌 𝖺 𝗉𝗋𝗈𝗉𝗈𝗌𝗂𝗍𝗂𝗈𝗇𝖺𝗅 𝖿𝗈𝗋𝗆𝗎𝗅𝖺 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖾𝖽 𝖿𝗋𝗈𝗆 𝗍𝗁𝖺𝗍 𝖺𝗑𝗂𝗈𝗆. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓐, 𝐏 ⊢ 𝐏), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (𝑟₁(𝑜₁, 𝑜₂) ⟺ 𝑟₂(𝑜₃)). ∎

𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑏𝑖𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙-𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛-2): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑏𝑖𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙-𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛-2 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐏₁ ⟺ 𝐐₁) ⊢ (𝐐₁ ⟹ 𝐏₁))⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.

𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂) - The proposition of interest: (𝑟₂(𝑜₃) ⟹ 𝑟₁(𝑜₁, 𝑜₂)). 𝗣𝗿𝗼𝗼𝗳: (𝑟₁(𝑜₁, 𝑜₂) ⟺ 𝑟₂(𝑜₃)), 𝗈𝖿 𝗍𝗁𝖾 𝖿𝗈𝗋𝗆 (𝐏₁ ⟺ 𝐐₁), 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁).  𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑏𝑖𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙-𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛-2 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₁ ⟺ 𝐐₁) ⊢ (𝐐₁ ⟹ 𝐏₁)), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 (𝑟₂(𝑜₃) ⟹ 𝑟₁(𝑜₁, 𝑜₂)). ∎

import punctilious as pu

# Create a universe-of-discourse with basic objects for the sake of this example.
u = pu.UniverseOfDiscourse(echo=True)
o1 = u.o.declare()
o2 = u.o.declare()
o3 = u.o.declare()
r1 = u.r.declare(2, signal_proposition=True)
r2 = u.r.declare(1, signal_proposition=True)
axiom = u.a.declare(natural_language='Dummy axiom for demonstration purposes')

# Elaborate a dummy theory with a set of propositions necessary for our demonstration
t1 = u.t(echo=True)
theory_axiom = t1.include_axiom(a=axiom)
phi1 = t1.i.axiom_interpretation.infer_formula_statement(a=theory_axiom,
    p=r1(o1, o2) | u.r.biconditional | r2(o3))

# And finally, use the biconditional-elimination-2 inference-rule:
proposition_of_interest = t1.i.biconditional_elimination_2.infer_formula_statement(p_iff_q=phi1,
    subtitle='The proposition of interest', echo=True)

Let "U7" be a universe-of-discourse.

Let "T1" be a theory-derivation in U7.

Axiom (T1.A1): Let axiom A1 "Dummy axiom for demonstration purposes" 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): (r1(o1, o2) <==> r2(o3)). Proof: "Dummy axiom for demonstration purposes" is postulated by axiom (A1). (r1(o1, o2) <==> r2(o3)) is a propositional formula interpreted from that axiom. Therefore, by the axiom-interpretation inference rule: (A, P |- P), it follows that (r1(o1, o2) <==> r2(o3)). QED

Inference rule (biconditional-elimination-2): Let inference-rule biconditional-elimination-2 defined as "((P1 <==> Q1) |- (Q1 ==> P1))" be included and considered valid in T1.

Proposition (T1.P2) - The proposition of interest: (r2(o3) ==> r1(o1, o2)). Proof: (r1(o1, o2) <==> r2(o3)), of the form (P1 <==> Q1), follows from prop. (P1).  Therefore, by the biconditional-elimination-2 inference rule: ((P1 <==> Q1) |- (Q1 ==> P1)), it follows that (r2(o3) ==> r1(o1, o2)). QED