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Theorem condandc 786
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 pm2.65 595, 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 → (𝜑𝜓))

Proof of Theorem condandc
StepHypRef Expression
1 condandc.1 . . 3 ((𝜑 ∧ ¬ 𝜓) → 𝜒)
2 condandc.2 . . 3 ((𝜑 ∧ ¬ 𝜓) → ¬ 𝜒)
31, 2pm2.65da 597 . 2 (𝜑 → ¬ ¬ 𝜓)
4 notnotrdc 762 . 2 (DECID 𝜓 → (¬ ¬ 𝜓𝜓))
53, 4syl5 32 1 (DECID 𝜓 → (𝜑𝜓))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 101  DECID wdc 753
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-in1 554  ax-in2 555  ax-io 640
This theorem depends on definitions:  df-bi 114  df-dc 754
This theorem is referenced by: (None)
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