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Theorem nfcdeq 3302
 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 2175 . 2
3 nfv 1754 . . 3
4 nfcdeq.2 . . . 4 CondEq
54cdeqri 3291 . . 3
63, 5sbie 2203 . 2
72, 6bitr3i 254 1
 Colors of variables: wff setvar class Syntax hints:   wb 187  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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