Theorem cdeqth 2802

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

Hypothesis

Ref	Expression
cdeqth.1	|- ph

Assertion

Ref	Expression
cdeqth	|- CondEq ( x = y -> ph )

Proof of Theorem cdeqth

Step	Hyp	Ref	Expression
1		cdeqth.1	. . 3 |- ph
2	1	a1i 9	. 2 |- ( x = y -> ph )
3	2	cdeqi 2800	1 |- CondEq ( x = y -> ph )

Syntax hints:  CondEqwcdeq 2798

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106

This theorem depends on definitions:  df-bi 115  df-cdeq 2799

This theorem is referenced by:  cdeqal1  2806  cdeqab1  2807  nfccdeq  2813