Theorem cdeqab1 2943

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
cdeqab1	|- CondEq ( x = y -> { x | ph } = { y | ps } )

Distinct variable groups:   ps, x   ph, y

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

Proof of Theorem cdeqab1

Step	Hyp	Ref	Expression
1		cdeqnot.1	. . . 4 |- CondEq ( x = y -> ( ph <-> ps ) )
2	1	cdeqri 2937	. . 3 |- ( x = y -> ( ph <-> ps ) )
3	2	cbvabv 2291	. 2 |- { x | ph } = { y | ps }
4	3	cdeqth 2938	1 |- CondEq ( x = y -> { x | ph } = { y | ps } )

Syntax hints:   <-> wb 104   = wceq 1343   {cab 2151  CondEqwcdeq 2934

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-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-cdeq 2935

This theorem is referenced by: (None)