Theorem cdeqel 2905

Description: Distribute conditional equality over elementhood. (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
cdeqel	|- CondEq ( x = y -> ( A e. C <-> B e. D ) )

Proof of Theorem cdeqel

Step	Hyp	Ref	Expression
1		cdeqeq.1	. . . 4 |- CondEq ( x = y -> A = B )
2	1	cdeqri 2895	. . 3 |- ( x = y -> A = B )
3		cdeqeq.2	. . . 4 |- CondEq ( x = y -> C = D )
4	3	cdeqri 2895	. . 3 |- ( x = y -> C = D )
5	2, 4	eleq12d 2210	. 2 |- ( x = y -> ( A e. C <-> B e. D ) )
6	5	cdeqi 2894	1 |- CondEq ( x = y -> ( A e. C <-> B e. D ) )

Syntax hints:   <-> wb 104   = wceq 1331   e. wcel 1480  CondEqwcdeq 2892

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-ie1 1469  ax-ie2 1470  ax-4 1487  ax-17 1506  ax-ial 1514  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-cleq 2132  df-clel 2135  df-cdeq 2893

This theorem is referenced by:  nfccdeq  2907