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Theorem cdeqab1 2996
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 . . . . 5  |- CondEq ( x  =  y  ->  ( ph 
<->  ps ) )
21cdeqri 2990 . . . 4  |-  ( x  =  y  ->  ( ph 
<->  ps ) )
32cbvabv 2415 . . 3  |-  { x  |  ph }  =  {
y  |  ps }
43a1i 10 . 2  |-  ( x  =  y  ->  { x  |  ph }  =  {
y  |  ps }
)
54cdeqi 2989 1  |- CondEq ( x  =  y  ->  { x  |  ph }  =  {
y  |  ps }
)
Colors of variables: wff set class
Syntax hints:    <-> wb 176    = wceq 1632   {cab 2282  CondEqwcdeq 2987
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1536  ax-5 1547  ax-17 1606  ax-9 1644  ax-8 1661  ax-6 1715  ax-7 1720  ax-11 1727  ax-12 1878  ax-ext 2277
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1310  df-ex 1532  df-nf 1535  df-sb 1639  df-clab 2283  df-cleq 2289  df-cdeq 2988
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