Theorem cdeqnot 3698

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 3696	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	notbid 317	. 2 ⊢ (𝑥 = 𝑦 → (¬ 𝜑 ↔ ¬ 𝜓))
4	3	cdeqi 3695	1 ⊢ CondEq(𝑥 = 𝑦 → (¬ 𝜑 ↔ ¬ 𝜓))

Syntax hints:  ¬ wn 3   ↔ wb 205  CondEqwcdeq 3693

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 206  df-cdeq 3694

This theorem is referenced by: (None)