Theorem cdeqnot 2752

Description: Distribute conditional equality over negation. (Contributed by Mario Carneiro, 11-Aug-2016.)

Hypothesis

Ref	Expression
cdeqnot.1	⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
cdeqnot	⊢ CondEq(𝑥 = 𝑦 → (¬ 𝜑 ↔ ¬ 𝜓))

Proof of Theorem cdeqnot

Step	Hyp	Ref	Expression
1		cdeqnot.1	. . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	cdeqri 2750	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	notbid 592	. 2 ⊢ (𝑥 = 𝑦 → (¬ 𝜑 ↔ ¬ 𝜓))
4	3	cdeqi 2749	1 ⊢ CondEq(𝑥 = 𝑦 → (¬ 𝜑 ↔ ¬ 𝜓))

Syntax hints:  ¬ wn 3   ↔ wb 98  CondEqwcdeq 2747

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545

This theorem depends on definitions:  df-bi 110  df-cdeq 2748

This theorem is referenced by: (None)