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Theorem nf5d 2118
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 2082 . 2 (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓))
4 nf5-1 2023 . 2 (∀𝑥(𝜓 → ∀𝑥𝜓) → Ⅎ𝑥𝜓)
53, 4syl 17 1 (𝜑 → Ⅎ𝑥𝜓)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wal 1481  wnf 1708
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047
This theorem depends on definitions:  df-bi 197  df-ex 1705  df-nf 1710
This theorem is referenced by:  axc16nfOLD  2163  nfaldOLD  2166  dvelimhw  2173  cbv1h  2268  nfeqf  2301  axc16nfALT  2323  nfsb2  2360  distel  31709  bj-cbv1hv  32730  wl-ax11-lem3  33364
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