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Theorem nfcdeq 3158
 Description: If we have a conditional equality proof, where is and is , and in fact does not have free in it according to , then unconditionally. This proves that is actually a not-free predicate. (Contributed by Mario Carneiro, 11-Aug-2016.)
Hypotheses
Ref Expression
nfcdeq.1
nfcdeq.2 CondEq
Assertion
Ref Expression
nfcdeq
Distinct variable groups:   ,   ,
Allowed substitution hints:   ()   ()

Proof of Theorem nfcdeq
StepHypRef Expression
1 nfcdeq.1 . . 3
21sbf 2117 . 2
3 nfv 1629 . . 3
4 nfcdeq.2 . . . 4 CondEq
54cdeqri 3147 . . 3
63, 5sbie 2149 . 2
72, 6bitr3i 243 1
 Colors of variables: wff set class Syntax hints:   wb 177  wnf 1553  wsb 1658  CondEqwcdeq 3144 This theorem is referenced by:  nfccdeq  3159 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1555  ax-5 1566  ax-17 1626  ax-9 1666  ax-8 1687  ax-6 1744  ax-7 1749  ax-11 1761  ax-12 1950 This theorem depends on definitions:  df-bi 178  df-or 360  df-an 361  df-tru 1328  df-ex 1551  df-nf 1554  df-sb 1659  df-cdeq 3145
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