Theorem cdeqi 2970

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 2969	. 2 ⊢ (CondEq(𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜑))
3	1, 2	mpbir 146	1 ⊢ CondEq(𝑥 = 𝑦 → 𝜑)

Syntax hints:   → wi 4  CondEqwcdeq 2968

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This theorem depends on definitions:  df-bi 117  df-cdeq 2969

This theorem is referenced by:  cdeqth  2972  cdeqnot  2973  cdeqal  2974  cdeqab  2975  cdeqim  2978  cdeqcv  2979  cdeqeq  2980  cdeqel  2981