Theorem cdeqnot 2939

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 2937	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	notbid 657	. 2 ⊢ (𝑥 = 𝑦 → (¬ 𝜑 ↔ ¬ 𝜓))
4	3	cdeqi 2936	1 ⊢ CondEq(𝑥 = 𝑦 → (¬ 𝜑 ↔ ¬ 𝜓))

Syntax hints:  ¬ wn 3   ↔ wb 104  CondEqwcdeq 2934

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605

This theorem depends on definitions:  df-bi 116  df-cdeq 2935

This theorem is referenced by: (None)