Theorem ifpn 1067

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

Step	Hyp	Ref	Expression
1		notnotb 317	. . . 4 ⊢ (𝜑 ↔ ¬ ¬ 𝜑)
2	1	imbi1i 352	. . 3 ⊢ ((𝜑 → 𝜓) ↔ (¬ ¬ 𝜑 → 𝜓))
3	2	anbi2ci 626	. 2 ⊢ (((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)) ↔ ((¬ 𝜑 → 𝜒) ∧ (¬ ¬ 𝜑 → 𝜓)))
4		dfifp2 1059	. 2 ⊢ (if-(𝜑, 𝜓, 𝜒) ↔ ((𝜑 → 𝜓) ∧ (¬ 𝜑 → 𝜒)))
5		dfifp2 1059	. 2 ⊢ (if-(¬ 𝜑, 𝜒, 𝜓) ↔ ((¬ 𝜑 → 𝜒) ∧ (¬ ¬ 𝜑 → 𝜓)))
6	3, 4, 5	3bitr4i 305	1 ⊢ (if-(𝜑, 𝜓, 𝜒) ↔ if-(¬ 𝜑, 𝜒, 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 398  if-wif 1057

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 209  df-an 399  df-or 844  df-ifp 1058

This theorem is referenced by:  ifpfal  1069  ifpdfbi  39832  ifpxorcor  39835