Theorem cdeqnot 3726

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 3724	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	notbid 321	. 2 ⊢ (𝑥 = 𝑦 → (¬ 𝜑 ↔ ¬ 𝜓))
4	3	cdeqi 3723	1 ⊢ CondEq(𝑥 = 𝑦 → (¬ 𝜑 ↔ ¬ 𝜓))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209  CondEqwcdeq 3721

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-cdeq 3722

This theorem is used by: (None)