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Theorem nfcdeq 3302
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 2175 . 2  |-  ( [ y  /  x ] ph 
<-> 
ph )
3 nfv 1754 . . 3  |-  F/ x ps
4 nfcdeq.2 . . . 4  |- CondEq ( x  =  y  ->  ( ph 
<->  ps ) )
54cdeqri 3291 . . 3  |-  ( x  =  y  ->  ( ph 
<->  ps ) )
63, 5sbie 2203 . 2  |-  ( [ y  /  x ] ph 
<->  ps )
72, 6bitr3i 254 1  |-  ( ph  <->  ps )
Colors of variables: wff setvar class
Syntax hints:    <-> wb 187   F/wnf 1663   [wsb 1789  CondEqwcdeq 3288
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1665  ax-4 1678  ax-5 1751  ax-6 1797  ax-7 1841  ax-10 1889  ax-12 1907  ax-13 2055
This theorem depends on definitions:  df-bi 188  df-or 371  df-an 372  df-ex 1660  df-nf 1664  df-sb 1790  df-cdeq 3289
This theorem is referenced by:  nfccdeq  3303
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