Theorem cdeqi 2749

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 2748	. 2 ⊢ (CondEq(𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜑))
3	1, 2	mpbir 134	1 ⊢ CondEq(𝑥 = 𝑦 → 𝜑)

Syntax hints:   → wi 4  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

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

This theorem is referenced by:  cdeqth  2751  cdeqnot  2752  cdeqal  2753  cdeqab  2754  cdeqim  2757  cdeqcv  2758  cdeqeq  2759  cdeqel  2760