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Theorem nfcdeq 3414
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 , then 𝜑(𝑥) ↔ 𝜑(𝑦) unconditionally. This proves that 𝑥𝜑 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 2379 . 2 ([𝑦 / 𝑥]𝜑𝜑)
3 nfv 1840 . . 3 𝑥𝜓
4 nfcdeq.2 . . . 4 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
54cdeqri 3403 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
63, 5sbie 2407 . 2 ([𝑦 / 𝑥]𝜑𝜓)
72, 6bitr3i 266 1 (𝜑𝜓)
Colors of variables: wff setvar class
Syntax hints:  wb 196  wnf 1705  [wsb 1877  CondEqwcdeq 3400
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-10 2016  ax-12 2044  ax-13 2245
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1702  df-nf 1707  df-sb 1878  df-cdeq 3401
This theorem is referenced by:  nfccdeq  3415
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