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Theorem nfcdeq 2821
Description: If we have a conditional equality proof, where  ph is  ph ( x ) and  ps is  ph (
y ), and  ph (
x ) in fact does not have  x free in it according to  F/, then  ph ( x )  <->  ph ( y ) unconditionally. This proves that  F/ x ph is actually a not-free predicate. (Contributed by Mario Carneiro, 11-Aug-2016.)
Hypotheses
Ref Expression
nfcdeq.1  |-  F/ x ph
nfcdeq.2  |- CondEq ( x  =  y  ->  ( ph 
<->  ps ) )
Assertion
Ref Expression
nfcdeq  |-  ( ph  <->  ps )
Distinct variable groups:    ps, x    ph, y
Allowed substitution hints:    ph( x)    ps( y)

Proof of Theorem nfcdeq
StepHypRef Expression
1 nfcdeq.1 . . 3  |-  F/ x ph
21sbf 1702 . 2  |-  ( [ y  /  x ] ph 
<-> 
ph )
3 nfv 1462 . . 3  |-  F/ x ps
4 nfcdeq.2 . . . 4  |- CondEq ( x  =  y  ->  ( ph 
<->  ps ) )
54cdeqri 2810 . . 3  |-  ( x  =  y  ->  ( ph 
<->  ps ) )
63, 5sbie 1716 . 2  |-  ( [ y  /  x ] ph 
<->  ps )
72, 6bitr3i 184 1  |-  ( ph  <->  ps )
Colors of variables: wff set class
Syntax hints:    <-> wb 103   F/wnf 1390   [wsb 1687  CondEqwcdeq 2807
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1377  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468
This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1688  df-cdeq 2808
This theorem is referenced by:  nfccdeq  2822
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