Theorem condandc 882

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 pm2.65 660, 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 662	. 2 ⊢ (𝜑 → ¬ ¬ 𝜓)
4		notnotrdc 844	. 2 ⊢ (DECID 𝜓 → (¬ ¬ 𝜓 → 𝜓))
5	3, 4	syl5 32	1 ⊢ (DECID 𝜓 → (𝜑 → 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104  DECID wdc 835

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710

This theorem depends on definitions:  df-bi 117  df-dc 836

This theorem is referenced by:  ifnetruedc  3602  perfectlem2  15236