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Theorem ifpn 1069
 Description: Conditional operator for the negation of a proposition. (Contributed by BJ, 30-Sep-2019.) (Proof shortened by Wolf Lammen, 5-May-2024.)
Assertion
Ref Expression
ifpn (if-(𝜑, 𝜓, 𝜒) ↔ if-(¬ 𝜑, 𝜒, 𝜓))

Proof of Theorem ifpn
StepHypRef Expression
1 ancom 464 . 2 (((¬ 𝜑𝜓) ∧ (¬ 𝜑𝜒)) ↔ ((¬ 𝜑𝜒) ∧ (¬ 𝜑𝜓)))
2 dfifp5 1063 . 2 (if-(𝜑, 𝜓, 𝜒) ↔ ((¬ 𝜑𝜓) ∧ (¬ 𝜑𝜒)))
3 dfifp3 1061 . 2 (if-(¬ 𝜑, 𝜒, 𝜓) ↔ ((¬ 𝜑𝜒) ∧ (¬ 𝜑𝜓)))
41, 2, 33bitr4i 306 1 (if-(𝜑, 𝜓, 𝜒) ↔ if-(¬ 𝜑, 𝜒, 𝜓))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 399   ∨ wo 844  if-wif 1058 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 210  df-an 400  df-or 845  df-ifp 1059 This theorem is referenced by:  ifpfal  1072  wl-ifp-ncond2  35033  wl-df3xor2  35037  wl-2xor  35051  wl-df3maxtru1  35060  ifpxorcor  40355
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