Theorem cdeqeq 2899

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

Hypotheses

Ref	Expression
cdeqeq.1	|- CondEq ( x = y -> A = B )
cdeqeq.2	|- CondEq ( x = y -> C = D )

Assertion

Ref	Expression
cdeqeq	|- CondEq ( x = y -> ( A = C <-> B = D ) )

Proof of Theorem cdeqeq

Step	Hyp	Ref	Expression
1		cdeqeq.1	. . . 4 |- CondEq ( x = y -> A = B )
2	1	cdeqri 2890	. . 3 |- ( x = y -> A = B )
3		cdeqeq.2	. . . 4 |- CondEq ( x = y -> C = D )
4	3	cdeqri 2890	. . 3 |- ( x = y -> C = D )
5	2, 4	eqeq12d 2152	. 2 |- ( x = y -> ( A = C <-> B = D ) )
6	5	cdeqi 2889	1 |- CondEq ( x = y -> ( A = C <-> B = D ) )

Syntax hints:   <-> wb 104   = wceq 1331  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  ax-5 1423  ax-gen 1425  ax-4 1487  ax-17 1506  ax-ext 2119

This theorem depends on definitions:  df-bi 116  df-cleq 2130  df-cdeq 2888

This theorem is referenced by: (None)