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Theorem nfdh 1524
Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)
Hypotheses
Ref Expression
nfdh.1 (𝜑 → ∀𝑥𝜑)
nfdh.2 (𝜑 → (𝜓 → ∀𝑥𝜓))
Assertion
Ref Expression
nfdh (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nfdh
StepHypRef Expression
1 nfdh.1 . . 3 (𝜑 → ∀𝑥𝜑)
21nfi 1462 . 2 𝑥𝜑
3 nfdh.2 . 2 (𝜑 → (𝜓 → ∀𝑥𝜓))
42, 3nfd 1523 1 (𝜑 → Ⅎ𝑥𝜓)
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1351  wnf 1460
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1447  ax-gen 1449  ax-4 1510
This theorem depends on definitions:  df-bi 117  df-nf 1461
This theorem is referenced by:  hbsbd  1982
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