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Theorem nfcdeq 3001
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 1979 . 2  |-  ( [ y  /  x ] ph 
<-> 
ph )
3 nfv 1609 . . 3  |-  F/ x ps
4 nfcdeq.2 . . . 4  |- CondEq ( x  =  y  ->  ( ph 
<->  ps ) )
54cdeqri 2990 . . 3  |-  ( x  =  y  ->  ( ph 
<->  ps ) )
63, 5sbie 1991 . 2  |-  ( [ y  /  x ] ph 
<->  ps )
72, 6bitr3i 242 1  |-  ( ph  <->  ps )
Colors of variables: wff set class
Syntax hints:    <-> wb 176   F/wnf 1534   [wsb 1638  CondEqwcdeq 2987
This theorem is referenced by:  nfccdeq  3002
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
This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1310  df-ex 1532  df-nf 1535  df-sb 1639  df-cdeq 2988
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