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Theorem bifald 32854
Description: Infer the equivalence to a contradiction from a negation, in deduction form. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
Hypothesis
Ref Expression
bifald.1 (𝜑 → ¬ 𝜓)
Assertion
Ref Expression
bifald (𝜑 → (𝜓 ↔ ⊥))

Proof of Theorem bifald
StepHypRef Expression
1 bifald.1 . 2 (𝜑 → ¬ 𝜓)
2 id 22 . . 3 (𝜓𝜓)
3 falim 1488 . . 3 (⊥ → 𝜓)
42, 3pm5.21ni 365 . 2 𝜓 → (𝜓 ↔ ⊥))
51, 4syl 17 1 (𝜑 → (𝜓 ↔ ⊥))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 194  wfal 1479
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8
This theorem depends on definitions:  df-bi 195  df-tru 1477  df-fal 1480
This theorem is referenced by: (None)
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