Theorem cdeqab 2945

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

Hypothesis

Ref	Expression
cdeqnot.1	|- CondEq ( x = y -> ( ph <-> ps ) )

Assertion

Ref	Expression
cdeqab	|- CondEq ( x = y -> { z | ph } = { z | ps } )

Distinct variable groups:   x, z   y, z

Allowed substitution hints:   ph( x, y, z)   ps( x, y, z)

Proof of Theorem cdeqab

Step	Hyp	Ref	Expression
1		cdeqnot.1	. . . 4 |- CondEq ( x = y -> ( ph <-> ps ) )
2	1	cdeqri 2941	. . 3 |- ( x = y -> ( ph <-> ps ) )
3	2	abbidv 2288	. 2 |- ( x = y -> { z | ph } = { z | ps } )
4	3	cdeqi 2940	1 |- CondEq ( x = y -> { z | ph } = { z | ps } )

Syntax hints:   <-> wb 104   = wceq 1348   {cab 2156  CondEqwcdeq 2938

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 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-11 1499  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-cdeq 2939

This theorem is referenced by: (None)