Theorem cdeqi 2889

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

Hypothesis

Ref	Expression
cdeqi.1	|- ( x = y -> ph )

Assertion

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

Proof of Theorem cdeqi

Step	Hyp	Ref	Expression
1		cdeqi.1	. 2 |- ( x = y -> ph )
2		df-cdeq 2888	. 2 |- (CondEq ( x = y -> ph ) <-> ( x = y -> ph ) )
3	1, 2	mpbir 145	1 |- CondEq ( x = y -> ph )

Syntax hints:   -> wi 4  CondEqwcdeq 2887

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116  df-cdeq 2888

This theorem is referenced by:  cdeqth  2891  cdeqnot  2892  cdeqal  2893  cdeqab  2894  cdeqim  2897  cdeqcv  2898  cdeqeq  2899  cdeqel  2900