Theorem cdeqab 3425

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 3421	. . 3 |- ( x = y -> ( ph <-> ps ) )
3	2	abbidv 2741	. 2 |- ( x = y -> { z | ph } = { z | ps } )
4	3	cdeqi 3420	1 |- CondEq ( x = y -> { z | ph } = { z | ps } )

Syntax hints:   <-> wb 196   = wceq 1483   {cab 2608  CondEqwcdeq 3418

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-cdeq 3419

This theorem is referenced by: (None)