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Theorem cdeqab1 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
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
StepHypRef Expression
1 cdeqnot.1 . . . 4  |- CondEq ( x  =  y  ->  ( ph 
<->  ps ) )
21cdeqri 3147 . . 3  |-  ( x  =  y  ->  ( ph 
<->  ps ) )
32cbvabv 2555 . 2  |-  { x  |  ph }  =  {
y  |  ps }
43cdeqth 3148 1  |- CondEq ( x  =  y  ->  { x  |  ph }  =  {
y  |  ps }
)
Colors of variables: wff set class
Syntax hints:    <-> wb 177    = wceq 1652   {cab 2422  CondEqwcdeq 3144
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1555  ax-5 1566  ax-17 1626  ax-9 1666  ax-8 1687  ax-6 1744  ax-7 1749  ax-11 1761  ax-12 1950  ax-ext 2417
This theorem depends on definitions:  df-bi 178  df-or 360  df-an 361  df-tru 1328  df-ex 1551  df-nf 1554  df-sb 1659  df-clab 2423  df-cleq 2429  df-cdeq 3145
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