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Theorem cdeqab 3153
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
StepHypRef Expression
1 cdeqnot.1 . . . 4  |- CondEq ( x  =  y  ->  ( ph 
<->  ps ) )
21cdeqri 3149 . . 3  |-  ( x  =  y  ->  ( ph 
<->  ps ) )
32abbidv 2552 . 2  |-  ( x  =  y  ->  { z  |  ph }  =  { z  |  ps } )
43cdeqi 3148 1  |- CondEq ( x  =  y  ->  { z  |  ph }  =  { z  |  ps } )
Colors of variables: wff set class
Syntax hints:    <-> wb 178    = wceq 1653   {cab 2424  CondEqwcdeq 3146
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1556  ax-5 1567  ax-17 1627  ax-9 1667  ax-8 1688  ax-6 1745  ax-7 1750  ax-11 1762  ax-12 1951  ax-ext 2419
This theorem depends on definitions:  df-bi 179  df-an 362  df-tru 1329  df-ex 1552  df-nf 1555  df-sb 1660  df-clab 2425  df-cleq 2431  df-cdeq 3147
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