Theorem cdeqi 3700

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

Hypothesis

Ref	Expression
cdeqi.1	⊢ (𝑥 = 𝑦 → 𝜑)

Assertion

Ref	Expression
cdeqi	⊢ CondEq(𝑥 = 𝑦 → 𝜑)

Proof of Theorem cdeqi

Step	Hyp	Ref	Expression
1		cdeqi.1	. 2 ⊢ (𝑥 = 𝑦 → 𝜑)
2		df-cdeq 3699	. 2 ⊢ (CondEq(𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜑))
3	1, 2	mpbir 230	1 ⊢ CondEq(𝑥 = 𝑦 → 𝜑)

Syntax hints:   → wi 4  CondEqwcdeq 3698

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 3699

This theorem is referenced by:  cdeqth  3702  cdeqnot  3703  cdeqal  3704  cdeqab  3705  cdeqim  3708  cdeqcv  3709  cdeqeq  3710  cdeqel  3711