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Theorem nf5d 2104
Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)
Hypotheses
Ref Expression
nf5d.1 𝑥𝜑
nf5d.2 (𝜑 → (𝜓 → ∀𝑥𝜓))
Assertion
Ref Expression
nf5d (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nf5d
StepHypRef Expression
1 nf5d.1 . . 3 𝑥𝜑
2 nf5d.2 . . 3 (𝜑 → (𝜓 → ∀𝑥𝜓))
31, 2alrimi 2069 . 2 (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓))
4 nf5-1 2010 . 2 (∀𝑥(𝜓 → ∀𝑥𝜓) → Ⅎ𝑥𝜓)
53, 4syl 17 1 (𝜑 → Ⅎ𝑥𝜓)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wal 1473  wnf 1699
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-12 2034
This theorem depends on definitions:  df-bi 196  df-ex 1696  df-nf 1701
This theorem is referenced by:  axc16nfOLD  2149  nfaldOLD  2152  dvelimhw  2159  cbv1h  2256  nfeqf  2289  axc16nfALT  2311  nfsb2  2348  distel  30953  bj-cbv1hv  31917  wl-ax11-lem3  32543
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