MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  nfcdeq Structured version   Visualization version   GIF version

Theorem nfcdeq 3716
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. Usage of this theorem is discouraged because it depends on ax-13 2374. (Contributed by Mario Carneiro, 11-Aug-2016.) (New usage is discouraged.)
Hypotheses
Ref Expression
nfcdeq.1 𝑥𝜑
nfcdeq.2 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
nfcdeq (𝜑𝜓)
Distinct variable group:   𝜓,𝑥
Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑦)

Proof of Theorem nfcdeq
StepHypRef Expression
1 nfcdeq.1 . . 3 𝑥𝜑
21sbf 2267 . 2 ([𝑦 / 𝑥]𝜑𝜑)
3 nfv 1921 . . 3 𝑥𝜓
4 nfcdeq.2 . . . 4 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
54cdeqri 3705 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
63, 5sbie 2508 . 2 ([𝑦 / 𝑥]𝜑𝜓)
72, 6bitr3i 276 1 (𝜑𝜓)
Colors of variables: wff setvar class
Syntax hints:  wb 205  wnf 1790  [wsb 2071  CondEqwcdeq 3702
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2015  ax-10 2141  ax-12 2175  ax-13 2374
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-ex 1787  df-nf 1791  df-sb 2072  df-cdeq 3703
This theorem is referenced by:  nfccdeq  3717
  Copyright terms: Public domain W3C validator