Theorem condandc 871

Description: Proof by contradiction. This only holds for decidable propositions, as it is part of the family of theorems which assume ¬ 𝜓, derive a contradiction, and therefore conclude 𝜓. By contrast, assuming 𝜓, deriving a contradiction, and therefore concluding ¬ 𝜓, as in pm2.65 649, is valid for all propositions. (Contributed by Jim Kingdon, 13-May-2018.)

Hypotheses

Ref	Expression
condandc.1	⊢ ((𝜑 ∧ ¬ 𝜓) → 𝜒)
condandc.2	⊢ ((𝜑 ∧ ¬ 𝜓) → ¬ 𝜒)

Assertion

Ref	Expression
condandc	⊢ (DECID 𝜓 → (𝜑 → 𝜓))

Proof of Theorem condandc

Step	Hyp	Ref	Expression
1		condandc.1	. . 3 ⊢ ((𝜑 ∧ ¬ 𝜓) → 𝜒)
2		condandc.2	. . 3 ⊢ ((𝜑 ∧ ¬ 𝜓) → ¬ 𝜒)
3	1, 2	pm2.65da 651	. 2 ⊢ (𝜑 → ¬ ¬ 𝜓)
4		notnotrdc 833	. 2 ⊢ (DECID 𝜓 → (¬ ¬ 𝜓 → 𝜓))
5	3, 4	syl5 32	1 ⊢ (DECID 𝜓 → (𝜑 → 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103  DECID wdc 824

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699

This theorem depends on definitions:  df-bi 116  df-dc 825

This theorem is referenced by: (None)