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Theorem nfcdeq 2761
 Description: If we have a conditional equality proof, where 𝜑 is 𝜑(𝑥) and 𝜓 is 𝜑(𝑦), and 𝜑(𝑥) in fact does not have 𝑥 free in it according to Ⅎ, then 𝜑(𝑥) ↔ 𝜑(𝑦) unconditionally. This proves that Ⅎ𝑥𝜑 is actually a not-free predicate. (Contributed by Mario Carneiro, 11-Aug-2016.)
Hypotheses
Ref Expression
nfcdeq.1 𝑥𝜑
nfcdeq.2 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
nfcdeq (𝜑𝜓)
Distinct variable groups:   𝜓,𝑥   𝜑,𝑦
Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)

Proof of Theorem nfcdeq
StepHypRef Expression
1 nfcdeq.1 . . 3 𝑥𝜑
21sbf 1660 . 2 ([𝑦 / 𝑥]𝜑𝜑)
3 nfv 1421 . . 3 𝑥𝜓
4 nfcdeq.2 . . . 4 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
54cdeqri 2750 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
63, 5sbie 1674 . 2 ([𝑦 / 𝑥]𝜑𝜓)
72, 6bitr3i 175 1 (𝜑𝜓)
 Colors of variables: wff set class Syntax hints:   ↔ wb 98  Ⅎwnf 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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