Theorem ifpn 1079

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 461	. 2 ⊢ (((¬ 𝜑 ∨ 𝜓) ∧ (¬ 𝜑 → 𝜒)) ↔ ((¬ 𝜑 → 𝜒) ∧ (¬ 𝜑 ∨ 𝜓)))
2		dfifp5 1073	. 2 ⊢ (if-(𝜑, 𝜓, 𝜒) ↔ ((¬ 𝜑 ∨ 𝜓) ∧ (¬ 𝜑 → 𝜒)))
3		dfifp3 1071	. 2 ⊢ (if-(¬ 𝜑, 𝜒, 𝜓) ↔ ((¬ 𝜑 → 𝜒) ∧ (¬ 𝜑 ∨ 𝜓)))
4	1, 2, 3	3bitr4i 304	1 ⊢ (if-(𝜑, 𝜓, 𝜒) ↔ if-(¬ 𝜑, 𝜒, 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 207   ∧ wa 396   ∨ wo 853  if-wif 1068

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 208  df-an 397  df-or 854  df-ifp 1069

This theorem is referenced by:  ifpfal  1081  wl-ifp-ncond2  37827  wl-df3xor2  37831  wl-2xor  37845  wl-df3maxtru1  37854  ifpxorcor  43920