Question

A natural deduction proofs using only 11 inference rules and no axioms

score 103 · Answer 1 · 2024-01-04 00:48:51

103

Views

A natural deduction proofs using only 11 inference rules and no axioms

Published on 04 Jan 2024 - 0:48

score 153 · Answer 2 · 2024-01-16 09:01:57

153

Views

Provided with the $\leftrightarrow$ elimination rule, is the substitution rule still neccessary in a natural deduction system?

Published on 16 Jan 2024 - 9:01

score 60 · Answer 3 · 2024-01-23 15:44:36

60

Views

Proving conjunction introduction in a natural deduction system with negation and disjunction

Published on 23 Jan 2024 - 15:44

score 171 · Answer 4 · 2026-02-23 12:08:07

171

Views

Proving $\exists x[P(x)] \to \exists y[\exists x[P(x)]\to P(y)]$ in for intuitionistic $\varepsilon$-calculus.

Published on 23 Feb 2026 - 12:08

score 116 · Answer 5 · 2024-02-14 00:27:20

116

Views

Help proving $\lnot$(G $\rightarrow$ $\lnot$A) $\vdash$ G.

Published on 14 Feb 2024 - 0:27

score 129 · Answer 6 · 2024-03-09 20:47:49

129

Views

Interpretation Theorem

Published on 09 Mar 2024 - 20:47

score 76 · Answer 7 · 2024-03-14 19:58:16

76

Views

Smallest natural deduction proof for Meredith's axiom from basic rules

Published on 14 Mar 2024 - 19:58

score 87 · Answer 8 · 2024-03-17 09:54:26

87

Views

Rationale behind Conditional Proof

Published on 17 Mar 2024 - 9:54

score 88 · Answer 9 · 2024-03-23 04:20:51

88

Views

Universal Quantification over Disjunction

Published on 23 Mar 2024 - 4:20

score 49 · Answer 10 · 2024-03-28 13:38:36

49

Views

Show the validity of ¬∀x(F x → Gx) ⊢ ∃xFx using natural deduction

Published on 28 Mar 2024 - 13:38

score 997 · Answer 11 · 2015-01-03 19:17:23

997

Views

Predicate natural deduction: Prove (∀x R(x,x)) => ∀x∃y R(x,y)

Published on 03 Jan 2015 - 19:17

score 386 · Answer 12 · 2015-01-23 20:16:03

386

Views

∀x(A→ B) ├ ¬∃x(A ∧ ¬B) prove with natural deduction

Published on 23 Jan 2015 - 20:16

score 836 · Answer 13 · 2015-01-27 16:10:18

836

Views

Show that $(p \to q) \lor (q \to p)$ is a tautology

Published on 27 Jan 2015 - 16:10

score 143 · Answer 14 · 2015-01-27 19:55:13

143

Views

how to prove $ B = ( \beta \Rightarrow \gamma) \Rightarrow (\alpha \Rightarrow \beta)\Rightarrow \alpha\Rightarrow \gamma $ using natural deduction

Published on 27 Jan 2015 - 19:55

score 1.2k · Answer 15 · 2015-01-28 20:02:30

1.2k

Views

$p \to (q\vee\neg r), \neg q, r ⊢ \neg p$ - Natural deduction- elimination with $\neg$ operator

Published on 28 Jan 2015 - 20:02

List Question

A natural deduction proofs using only 11 inference rules and no axioms

Provided with the $\leftrightarrow$ elimination rule, is the substitution rule still neccessary in a natural deduction system?

Proving conjunction introduction in a natural deduction system with negation and disjunction

Proving $\exists x[P(x)] \to \exists y[\exists x[P(x)]\to P(y)]$ in for intuitionistic $\varepsilon$-calculus.

Help proving $\lnot$(G $\rightarrow$ $\lnot$A) $\vdash$ G.

Interpretation Theorem

Smallest natural deduction proof for Meredith's axiom from basic rules

Rationale behind Conditional Proof

Universal Quantification over Disjunction

Show the validity of ¬∀x(F x → Gx) ⊢ ∃xFx using natural deduction

Predicate natural deduction: Prove (∀x R(x,x)) => ∀x∃y R(x,y)

∀x(A→ B) ├ ¬∃x(A ∧ ¬B) prove with natural deduction

Show that $(p \to q) \lor (q \to p)$ is a tautology

how to prove $ B = ( \beta \Rightarrow \gamma) \Rightarrow (\alpha \Rightarrow \beta)\Rightarrow \alpha\Rightarrow \gamma $ using natural deduction

$p \to (q\vee\neg r), \neg q, r ⊢ \neg p$ - Natural deduction- elimination with $\neg$ operator

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