ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  nfcdeq GIF version

Theorem nfcdeq 2813
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 1701 . 2 ([𝑦 / 𝑥]𝜑𝜑)
3 nfv 1462 . . 3 𝑥𝜓
4 nfcdeq.2 . . . 4 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
54cdeqri 2802 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
63, 5sbie 1715 . 2 ([𝑦 / 𝑥]𝜑𝜓)
72, 6bitr3i 184 1 (𝜑𝜓)
Colors of variables: wff set class
Syntax hints:  wb 103  wnf 1390  [wsb 1686  CondEqwcdeq 2799
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1377  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468
This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1687  df-cdeq 2800
This theorem is referenced by:  nfccdeq  2814
  Copyright terms: Public domain W3C validator