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Theorem nfdh 1433
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 1367 . 2 𝑥𝜑
3 nfdh.2 . 2 (𝜑 → (𝜓 → ∀𝑥𝜓))
42, 3nfd 1432 1 (𝜑 → Ⅎ𝑥𝜓)
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1257  wnf 1365
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-4 1416
This theorem depends on definitions:  df-bi 114  df-nf 1366
This theorem is referenced by:  hbsbd  1874
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