inconsistency-introduction-3 (python sample)๏
See also
math concept | python declaration class | python inclusion class
This page shows how to infer new statements in a theory-derivation by applying the inconsistency-introduction-3 inference-rule.
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.inconsistency_introduction_3.infer_statement(...)
Sample code๏
Code output๏
๐ซ๐พ๐ โ๐ฐโ
โโ ๐ป๐พ ๐บ ๐ข๐๐๐ฃ๐๐๐ ๐-๐๐-๐๐๐ ๐๐๐ข๐๐ ๐.
๐ซ๐พ๐ โ๐ฏโโ ๐ป๐พ ๐บ ๐กโ๐๐๐๐ฆ-๐๐๐๐๐ฃ๐๐ก๐๐๐ ๐๐ ๐ฐโ
โ.
๐๐
๐ถ๐ผ๐บ (๐ฏโ.๐ดโ): ๐ซ๐พ๐ ๐๐ฅ๐๐๐ ๐โ โ๐๐ถ๐ฎ๐ฎ๐บ ๐ข๐น๐ช๐ฐ๐ฎ ๐ง๐ฐ๐ณ ๐ฅ๐ฆ๐ฎ๐ฐ๐ฏ๐ด๐ต๐ณ๐ข๐ต๐ช๐ฐ๐ฏ ๐ฑ๐ถ๐ณ๐ฑ๐ฐ๐ด๐ฆ๐ดโ ๐ป๐พ ๐๐๐ผ๐
๐๐ฝ๐พ๐ฝ (๐๐๐๐๐๐
๐บ๐๐พ๐ฝ) ๐๐ ๐ฏโ.
๐๐ป๐ณ๐ฒ๐ฟ๐ฒ๐ป๐ฐ๐ฒ ๐ฟ๐๐น๐ฒ (๐๐ฅ๐๐๐-๐๐๐ก๐๐๐๐๐๐ก๐๐ก๐๐๐): ๐ซ๐พ๐ ๐๐๐๐๐๐๐๐๐-๐๐ข๐๐ ๐๐ฅ๐๐๐-๐๐๐ก๐๐๐๐๐๐ก๐๐ก๐๐๐ ๐ฝ๐พ๐ฟ๐๐๐พ๐ฝ ๐บ๐ โ(๐, ๐ โข ๐)โ ๐ป๐พ ๐๐๐ผ๐
๐๐ฝ๐พ๐ฝ ๐บ๐๐ฝ ๐ผ๐๐๐๐๐ฝ๐พ๐๐พ๐ฝ ๐๐บ๐
๐๐ฝ ๐๐ ๐ฏโ.
๐ฃ๐ฟ๐ผ๐ฝ๐ผ๐๐ถ๐๐ถ๐ผ๐ป (๐ฏโ.๐โ): (๐โ โ ๐โ). ๐ฃ๐ฟ๐ผ๐ผ๐ณ: โ๐๐ถ๐ฎ๐ฎ๐บ ๐ข๐น๐ช๐ฐ๐ฎ ๐ง๐ฐ๐ณ ๐ฅ๐ฆ๐ฎ๐ฐ๐ฏ๐ด๐ต๐ณ๐ข๐ต๐ช๐ฐ๐ฏ ๐ฑ๐ถ๐ณ๐ฑ๐ฐ๐ด๐ฆ๐ดโ ๐๐ ๐๐๐๐๐๐
๐บ๐๐พ๐ฝ ๐ป๐ ๐ฎ๐
๐ถ๐ผ๐บ (๐ดโ). (๐โ โ ๐โ) ๐๐ ๐บ ๐๐๐๐๐๐๐๐๐๐๐๐บ๐
๐ฟ๐๐๐๐๐
๐บ ๐๐๐๐พ๐๐๐๐พ๐๐พ๐ฝ ๐ฟ๐๐๐ ๐๐๐บ๐ ๐บ๐๐๐๐. ๐ณ๐๐พ๐๐พ๐ฟ๐๐๐พ, ๐ป๐ ๐๐๐พ ๐๐ฅ๐๐๐-๐๐๐ก๐๐๐๐๐๐ก๐๐ก๐๐๐ ๐๐๐ฟ๐พ๐๐พ๐๐ผ๐พ ๐๐๐
๐พ: (๐, ๐ โข ๐), ๐๐ ๐ฟ๐๐
๐
๐๐๐ ๐๐๐บ๐ (๐โ โ ๐โ). โ
๐ซ๐พ๐ โ๐ฏโโ ๐ป๐พ ๐บ ๐กโ๐๐๐๐ฆ-๐๐๐๐๐ฃ๐๐ก๐๐๐ ๐๐ ๐ฐโ
โ.
๐๐ป๐ณ๐ฒ๐ฟ๐ฒ๐ป๐ฐ๐ฒ ๐ฟ๐๐น๐ฒ (๐๐๐๐๐๐ ๐๐ ๐ก๐๐๐๐ฆ-๐๐๐ก๐๐๐๐ข๐๐ก๐๐๐-3): ๐ซ๐พ๐ ๐๐๐๐๐๐๐๐๐-๐๐ข๐๐ ๐๐๐๐๐๐ ๐๐ ๐ก๐๐๐๐ฆ-๐๐๐ก๐๐๐๐ข๐๐ก๐๐๐-3 ๐ฝ๐พ๐ฟ๐๐๐พ๐ฝ ๐บ๐ โ((๐โ โ ๐โ) โข ๐ผ๐๐(๐ฏโ))โ ๐ป๐พ ๐๐๐ผ๐
๐๐ฝ๐พ๐ฝ ๐บ๐๐ฝ ๐ผ๐๐๐๐๐ฝ๐พ๐๐พ๐ฝ ๐๐บ๐
๐๐ฝ ๐๐ ๐ฏโ.
๐ฃ๐ฟ๐ผ๐ฝ๐ผ๐๐ถ๐๐ถ๐ผ๐ป (๐ฏโ.๐โ) - The proposition of interest: ๐ผ๐๐(๐ฏโ). ๐ฃ๐ฟ๐ผ๐ผ๐ณ: ๐ซ๐พ๐ (๐ท โ ๐ท) := (๐โ โ ๐โ), ๐๐๐๐ผ๐ ๐ฟ๐๐
๐
๐๐๐ ๐ฟ๐๐๐ ๐ฝ๐ฟ๐ผ๐ฝ. (๐โ). ๐ณ๐๐พ๐๐พ๐ฟ๐๐๐พ, ๐ป๐ ๐๐๐พ ๐๐๐๐๐๐ ๐๐ ๐ก๐๐๐๐ฆ-๐๐๐ก๐๐๐๐ข๐๐ก๐๐๐-3 ๐๐๐ฟ๐พ๐๐พ๐๐ผ๐พ ๐๐๐
๐พ: ((๐โ โ ๐โ) โข ๐ผ๐๐(๐ฏโ)), ๐๐ ๐ฟ๐๐
๐
๐๐๐ ๐๐๐บ๐ ๐ผ๐๐(๐ฏโ). โ
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()
t1 = u.t(echo=True)
axiom = u.a.declare(natural_language='Dummy axiom for demonstration purposes')
# Elaborate a dummy theory with inconsistent propositions
theory_axiom = t1.include_axiom(axiom)
x_unequal_x = t1.i.axiom_interpretation.infer_formula_statement(theory_axiom,
(o1 | u.r.unequal | o1))
t1.stabilize()
# Use a distinct theory T2 to demonstrate the inconsistency of T1
# because T1 could not prove its own inconsistency because it is inconsistent!
t2 = u.t(echo=True)
# And finally, use the inconsistency-introduction-3 inference-rule:
proposition_of_interest = t2.i.inconsistency_introduction_3.infer_formula_statement(
x_unequal_x=x_unequal_x, t=t1, subtitle='The proposition of interest')
Let "U57" be a universe-of-discourse.
Let "T1" be a theory-derivation in U57.
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): (o1 neq o1). Proof: "Dummy axiom for demonstration purposes" is postulated by axiom (A1). (o1 neq o1) is a propositional formula interpreted from that axiom. Therefore, by the axiom-interpretation inference rule: (A, P |- P), it follows that (o1 neq o1). QED
Let "T2" be a theory-derivation in U57.
Inference rule (inconsistency-introduction-3): Let inference-rule inconsistency-introduction-3 defined as "((P1 neq P1) |- Inc(T1))" be included and considered valid in T2.
Proposition (T2.P2) - The proposition of interest: Inc(T1). Proof: Let (P neq P) := (o1 neq o1), which follows from prop. (P1). Therefore, by the inconsistency-introduction-3 inference rule: ((P1 neq P1) |- Inc(T1)), it follows that Inc(T1). QED
Will be provided in a future version.
Will be provided in a future version.