Theorem had0OLD 1636

Description: Obsolete version of had0 1634 as of 10-Aug-2026. (Contributed by Mario Carneiro, 4-Sep-2016.) (Proof shortened by Wolf Lammen, 12-Jul-2020.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
had0OLD	⊢ (¬ 𝜑 → (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ⊻ 𝜒)))

Proof of Theorem had0OLD

Step	Hyp	Ref	Expression
1		had1OLD 1635	. . 3 ⊢ (¬ 𝜑 → (hadd(¬ 𝜑, ¬ 𝜓, ¬ 𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒)))
2		hadnot 1632	. . 3 ⊢ (¬ hadd(𝜑, 𝜓, 𝜒) ↔ hadd(¬ 𝜑, ¬ 𝜓, ¬ 𝜒))
3		xnor 1543	. . . 4 ⊢ ((𝜓 ↔ 𝜒) ↔ ¬ (𝜓 ⊻ 𝜒))
4		notbi 322	. . . 4 ⊢ ((𝜓 ↔ 𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒))
5	3, 4	bitr3i 280	. . 3 ⊢ (¬ (𝜓 ⊻ 𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒))
6	1, 2, 5	3bitr4g 317	. 2 ⊢ (¬ 𝜑 → (¬ hadd(𝜑, 𝜓, 𝜒) ↔ ¬ (𝜓 ⊻ 𝜒)))
7	6	con4bid 320	1 ⊢ (¬ 𝜑 → (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ⊻ 𝜒)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ⊻ wxo 1541  haddwhad 1623

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-xor 1542  df-had 1624

This theorem is used by:  hadifpOLD  1638