modus-ponens (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.modus_ponens.infer_statement(...)

Sample code

Code output

𝖫𝖾𝗍 ⌜𝒰₆₀⌝ 𝖻𝖾 𝖺 𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒-𝑜𝑓-𝑑𝑖𝑠𝑐𝑜𝑢𝑟𝑠𝑒.

𝖫𝖾𝗍 ⌜𝒯₁⌝ 𝖻𝖾 𝖺 𝑡ℎ𝑒𝑜𝑟𝑦-𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑜𝑛 𝗂𝗇 𝒰₆₀.

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)). ∎

𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₂): 𝑟₁(𝑜₁, 𝑜₂). 𝗣𝗿𝗼𝗼𝗳: ⌜𝘋𝘶𝘮𝘮𝘺 𝘢𝘹𝘪𝘰𝘮 𝘧𝘰𝘳 𝘥𝘦𝘮𝘰𝘯𝘴𝘵𝘳𝘢𝘵𝘪𝘰𝘯 𝘱𝘶𝘳𝘱𝘰𝘴𝘦𝘴⌝ 𝗂𝗌 𝗉𝗈𝗌𝗍𝗎𝗅𝖺𝗍𝖾𝖽 𝖻𝗒 𝗮𝘅𝗶𝗼𝗺 (𝐴₁). 𝑟₁(𝑜₁, 𝑜₂) 𝗂𝗌 𝖺 𝗉𝗋𝗈𝗉𝗈𝗌𝗂𝗍𝗂𝗈𝗇𝖺𝗅 𝖿𝗈𝗋𝗆𝗎𝗅𝖺 𝗂𝗇𝗍𝖾𝗋𝗉𝗋𝖾𝗍𝖾𝖽 𝖿𝗋𝗈𝗆 𝗍𝗁𝖺𝗍 𝖺𝗑𝗂𝗈𝗆. 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑎𝑥𝑖𝑜𝑚-𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡𝑎𝑡𝑖𝑜𝑛 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: (𝓐, 𝐏 ⊢ 𝐏), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 𝑟₁(𝑜₁, 𝑜₂). ∎

𝗜𝗻𝗳𝗲𝗿𝗲𝗻𝗰𝗲 𝗿𝘂𝗹𝗲 (𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠): 𝖫𝖾𝗍 𝑖𝑛𝑓𝑒𝑟𝑒𝑛𝑐𝑒-𝑟𝑢𝑙𝑒 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝖽𝖾𝖿𝗂𝗇𝖾𝖽 𝖺𝗌 ⌜((𝐏₁ ⟹ 𝐐₁), 𝐏₁ ⊢ 𝐐₁)⌝ 𝖻𝖾 𝗂𝗇𝖼𝗅𝗎𝖽𝖾𝖽 𝖺𝗇𝖽 𝖼𝗈𝗇𝗌𝗂𝖽𝖾𝗋𝖾𝖽 𝗏𝖺𝗅𝗂𝖽 𝗂𝗇 𝒯₁.

𝗣𝗿𝗼𝗽𝗼𝘀𝗶𝘁𝗶𝗼𝗻 (𝒯₁.𝑃₃) - The proposition of interest: 𝑟₂(𝑜₃). 𝗣𝗿𝗼𝗼𝗳: (𝑟₁(𝑜₁, 𝑜₂) ⟹ 𝑟₂(𝑜₃)) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₁).𝑟₁(𝑜₁, 𝑜₂) 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝖿𝗋𝗈𝗆 𝗽𝗿𝗼𝗽. (𝑃₂). 𝖳𝗁𝖾𝗋𝖾𝖿𝗈𝗋𝖾, 𝖻𝗒 𝗍𝗁𝖾 𝑚𝑜𝑑𝑢𝑠-𝑝𝑜𝑛𝑒𝑛𝑠 𝗂𝗇𝖿𝖾𝗋𝖾𝗇𝖼𝖾 𝗋𝗎𝗅𝖾: ((𝐏₁ ⟹ 𝐐₁), 𝐏₁ ⊢ 𝐐₁), 𝗂𝗍 𝖿𝗈𝗅𝗅𝗈𝗐𝗌 𝗍𝗁𝖺𝗍 𝑟₂(𝑜₃). ∎

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(arity=2, signal_proposition=True)
r2 = u.r.declare(arity=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)
p_implies_q = t1.i.axiom_interpretation.infer_formula_statement(a=theory_axiom,
    p=r1(o1, o2) | u.r.implies | r2(o3), lock=False)
p = t1.i.axiom_interpretation.infer_formula_statement(a=theory_axiom, p=r1(o1, o2), lock=True)

# And finally, use the modus-ponens inference-rule:
proposition_of_interest = t1.i.modus_ponens.infer_formula_statement(p_implies_q=p_implies_q, p=p,
    subtitle='The proposition of interest')

Let "U59" be a universe-of-discourse.

Let "T1" be a theory-derivation in U59.

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

Proposition (T1.P2): r1(o1, o2). Proof: "Dummy axiom for demonstration purposes" is postulated by axiom (A1). r1(o1, o2) is a propositional formula interpreted from that axiom. Therefore, by the axiom-interpretation inference rule: (A, P |- P), it follows that r1(o1, o2). QED

Inference rule (modus-ponens): Let inference-rule modus-ponens defined as "((P1 ==> Q1), P1 |- Q1)" be included and considered valid in T1.

Proposition (T1.P3) - The proposition of interest: r2(o3). Proof: (r1(o1, o2) ==> r2(o3)) follows from prop. (P1).r1(o1, o2) follows from prop. (P2). Therefore, by the modus-ponens inference rule: ((P1 ==> Q1), P1 |- Q1), it follows that r2(o3). QED