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Theorem fnopabeqd 26488
Description: Equality deduction for function abstractions. (Contributed by Jeff Madsen, 19-Jun-2011.)
Hypothesis
Ref Expression
fnopabeqd.1  |-  ( ph  ->  B  =  C )
Assertion
Ref Expression
fnopabeqd  |-  ( ph  ->  { <. x ,  y
>.  |  ( x  e.  A  /\  y  =  B ) }  =  { <. x ,  y
>.  |  ( x  e.  A  /\  y  =  C ) } )
Distinct variable groups:    ph, x    ph, y
Allowed substitution hints:    A( x, y)    B( x, y)    C( x, y)

Proof of Theorem fnopabeqd
StepHypRef Expression
1 fnopabeqd.1 . . . 4  |-  ( ph  ->  B  =  C )
21eqeq2d 2307 . . 3  |-  ( ph  ->  ( y  =  B  <-> 
y  =  C ) )
32anbi2d 684 . 2  |-  ( ph  ->  ( ( x  e.  A  /\  y  =  B )  <->  ( x  e.  A  /\  y  =  C ) ) )
43opabbidv 4098 1  |-  ( ph  ->  { <. x ,  y
>.  |  ( x  e.  A  /\  y  =  B ) }  =  { <. x ,  y
>.  |  ( x  e.  A  /\  y  =  C ) } )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 358    = wceq 1632    e. wcel 1696   {copab 4092
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-an 360  df-tru 1310  df-ex 1532  df-nf 1535  df-sb 1639  df-clab 2283  df-cleq 2289  df-opab 4094
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