Theorem cdeqab1 2952

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 2946	. . 3 |- ( x = y -> ( ph <-> ps ) )
3	2	cbvabv 2300	. 2 |- { x | ph } = { y | ps }
4	3	cdeqth 2947	1 |- CondEq ( x = y -> { x | ph } = { y | ps } )

Syntax hints:   <-> wb 105   = wceq 1353   {cab 2161  CondEqwcdeq 2943

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1445  ax-7 1446  ax-gen 1447  ax-ie1 1491  ax-ie2 1492  ax-8 1502  ax-10 1503  ax-11 1504  ax-i12 1505  ax-bndl 1507  ax-4 1508  ax-17 1524  ax-i9 1528  ax-ial 1532  ax-i5r 1533  ax-ext 2157

This theorem depends on definitions:  df-bi 117  df-nf 1459  df-sb 1761  df-clab 2162  df-cleq 2168  df-cdeq 2944

This theorem is referenced by: (None)