Theorem cdeqi 2801

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 2800	. 2 |- (CondEq ( x = y -> ph ) <-> ( x = y -> ph ) )
3	1, 2	mpbir 144	1 |- CondEq ( x = y -> ph )

Syntax hints:   -> wi 4  CondEqwcdeq 2799

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 2800

This theorem is referenced by:  cdeqth  2803  cdeqnot  2804  cdeqal  2805  cdeqab  2806  cdeqim  2809  cdeqcv  2810  cdeqeq  2811  cdeqel  2812