Theorem cdeqi 3420

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 3419	. 2 ⊢ (CondEq(𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜑))
3	1, 2	mpbir 221	1 ⊢ CondEq(𝑥 = 𝑦 → 𝜑)

Syntax hints:   → wi 4  CondEqwcdeq 3418

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 197  df-cdeq 3419

This theorem is referenced by:  cdeqth  3422  cdeqnot  3423  cdeqal  3424  cdeqab  3425  cdeqim  3428  cdeqcv  3429  cdeqeq  3430  cdeqel  3431  bj-cdeqab  32787