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Theorem ifpn 1016
 Description: Conditional operator for the negation of a proposition. (Contributed by BJ, 30-Sep-2019.)
Assertion
Ref Expression
ifpn (if-(𝜑, 𝜓, 𝜒) ↔ if-(¬ 𝜑, 𝜒, 𝜓))

Proof of Theorem ifpn
StepHypRef Expression
1 notnotb 303 . . . 4 (𝜑 ↔ ¬ ¬ 𝜑)
21imbi1i 338 . . 3 ((𝜑𝜓) ↔ (¬ ¬ 𝜑𝜓))
32anbi2ci 728 . 2 (((𝜑𝜓) ∧ (¬ 𝜑𝜒)) ↔ ((¬ 𝜑𝜒) ∧ (¬ ¬ 𝜑𝜓)))
4 dfifp2 1008 . 2 (if-(𝜑, 𝜓, 𝜒) ↔ ((𝜑𝜓) ∧ (¬ 𝜑𝜒)))
5 dfifp2 1008 . 2 (if-(¬ 𝜑, 𝜒, 𝜓) ↔ ((¬ 𝜑𝜒) ∧ (¬ ¬ 𝜑𝜓)))
63, 4, 53bitr4i 291 1 (if-(𝜑, 𝜓, 𝜒) ↔ if-(¬ 𝜑, 𝜒, 𝜓))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195   ∧ wa 383  if-wif 1006 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 196  df-or 384  df-an 385  df-ifp 1007 This theorem is referenced by:  ifpfal  1018  ifpdfbi  36837  ifpxorcor  36840
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