Theorem ifpn 1090

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 466	. 2 ⊢ (((¬ 𝜑 ∨ 𝜓) ∧ (¬ 𝜑 → 𝜒)) ↔ ((¬ 𝜑 → 𝜒) ∧ (¬ 𝜑 ∨ 𝜓)))
2		dfifp5 1083	. 2 ⊢ (if-(𝜑, 𝜓, 𝜒) ↔ ((¬ 𝜑 ∨ 𝜓) ∧ (¬ 𝜑 → 𝜒)))
3		dfifp3 1081	. 2 ⊢ (if-(¬ 𝜑, 𝜒, 𝜓) ↔ ((¬ 𝜑 → 𝜒) ∧ (¬ 𝜑 ∨ 𝜓)))
4	1, 2, 3	3bitr4i 306	1 ⊢ (if-(𝜑, 𝜓, 𝜒) ↔ if-(¬ 𝜑, 𝜒, 𝜓))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861  if-wif 1078

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ifp 1079

This theorem is used by:  ifpfal  1092  wl-ifp-ncond2  38144  wl-df3xor2  38148  wl-2xor  38162  wl-df3maxtru1  38171  ifpxorcor  44235