Theorem ifpn 1070

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

Step	Hyp	Ref	Expression
1		ancom 460	. 2 ⊢ (((¬ 𝜑 ∨ 𝜓) ∧ (¬ 𝜑 → 𝜒)) ↔ ((¬ 𝜑 → 𝜒) ∧ (¬ 𝜑 ∨ 𝜓)))
2		dfifp5 1064	. 2 ⊢ (if-(𝜑, 𝜓, 𝜒) ↔ ((¬ 𝜑 ∨ 𝜓) ∧ (¬ 𝜑 → 𝜒)))
3		dfifp3 1062	. 2 ⊢ (if-(¬ 𝜑, 𝜒, 𝜓) ↔ ((¬ 𝜑 → 𝜒) ∧ (¬ 𝜑 ∨ 𝜓)))
4	1, 2, 3	3bitr4i 302	1 ⊢ (if-(𝜑, 𝜓, 𝜒) ↔ if-(¬ 𝜑, 𝜒, 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 395   ∨ wo 843  if-wif 1059

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 206  df-an 396  df-or 844  df-ifp 1060

This theorem is referenced by:  ifpfal  1073  wl-ifp-ncond2  35563  wl-df3xor2  35567  wl-2xor  35581  wl-df3maxtru1  35590  ifpxorcor  40981