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Theorem nfd 1486
Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)
Hypotheses
Ref Expression
nfd.1 𝑥𝜑
nfd.2 (𝜑 → (𝜓 → ∀𝑥𝜓))
Assertion
Ref Expression
nfd (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nfd
StepHypRef Expression
1 nfd.1 . . . 4 𝑥𝜑
21nfri 1482 . . 3 (𝜑 → ∀𝑥𝜑)
3 nfd.2 . . 3 (𝜑 → (𝜓 → ∀𝑥𝜓))
42, 3alrimih 1428 . 2 (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓))
5 df-nf 1420 . 2 (Ⅎ𝑥𝜓 ↔ ∀𝑥(𝜓 → ∀𝑥𝜓))
64, 5sylibr 133 1 (𝜑 → Ⅎ𝑥𝜓)
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1312  wnf 1419
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 1406  ax-gen 1408  ax-4 1470
This theorem depends on definitions:  df-bi 116  df-nf 1420
This theorem is referenced by:  nfdh  1487  nfrimi  1488  nfnt  1617  cbv1h  1706  nfald  1716  a16nf  1820  dvelimALT  1961  dvelimfv  1962  nfsb4t  1965  hbeud  1997
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