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Theorem nfcdeq 2879
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 1735 . 2 ([𝑦 / 𝑥]𝜑𝜑)
3 nfv 1493 . . 3 𝑥𝜓
4 nfcdeq.2 . . . 4 CondEq(𝑥 = 𝑦 → (𝜑𝜓))
54cdeqri 2868 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
63, 5sbie 1749 . 2 ([𝑦 / 𝑥]𝜑𝜓)
72, 6bitr3i 185 1 (𝜑𝜓)
Colors of variables: wff set class
Syntax hints:  wb 104  wnf 1421  [wsb 1720  CondEqwcdeq 2865
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1408  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499
This theorem depends on definitions:  df-bi 116  df-nf 1422  df-sb 1721  df-cdeq 2866
This theorem is referenced by:  nfccdeq  2880
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