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Theorem nfcdeq 2761
 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 1660 . 2
3 nfv 1421 . . 3
4 nfcdeq.2 . . . 4 CondEq
54cdeqri 2750 . . 3
63, 5sbie 1674 . 2
72, 6bitr3i 175 1
 Colors of variables: wff set class Syntax hints:   wb 98  wnf 1349  wsb 1645  CondEqwcdeq 2747 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427 This theorem depends on definitions:  df-bi 110  df-nf 1350  df-sb 1646  df-cdeq 2748 This theorem is referenced by:  nfccdeq  2762
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