Theorem cdeqth 3729

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

Hypothesis

Ref	Expression
cdeqth.1	⊢ 𝜑

Assertion

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

Proof of Theorem cdeqth

Step	Hyp	Ref	Expression
1		cdeqth.1	. . 3 ⊢ 𝜑
2	1	a1i 11	. 2 ⊢ (𝑥 = 𝑦 → 𝜑)
3	2	cdeqi 3727	1 ⊢ CondEq(𝑥 = 𝑦 → 𝜑)

This proof depends on syntax axioms:  CondEqwcdeq 3725

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 3726

This theorem is used by:  cdeqal1  3733  cdeqab1  3734  nfccdeq  3740