Theorem cdeqab 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
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 2948	. . 3 |- ( x = y -> ( ph <-> ps ) )
3	2	abbidv 2295	. 2 |- ( x = y -> { z | ph } = { z | ps } )
4	3	cdeqi 2947	1 |- CondEq ( x = y -> { z | ph } = { z | ps } )

Syntax hints:   <-> wb 105   = wceq 1353   {cab 2163  CondEqwcdeq 2945

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-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-11 1506  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-cdeq 2946

This theorem is referenced by: (None)