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Theorem nf5dh 2200
 Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Mario Carneiro, 24-Sep-2016.) df-nf 1885 changed. (Revised by Wolf Lammen, 11-Oct-2021.)
Hypotheses
Ref Expression
nf5dh.1 (𝜑 → ∀𝑥𝜑)
nf5dh.2 (𝜑 → (𝜓 → ∀𝑥𝜓))
Assertion
Ref Expression
nf5dh (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nf5dh
StepHypRef Expression
1 nf5dh.1 . . 3 (𝜑 → ∀𝑥𝜑)
2 nf5dh.2 . . 3 (𝜑 → (𝜓 → ∀𝑥𝜓))
31, 2alrimih 1924 . 2 (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓))
4 nf5-1 2198 . 2 (∀𝑥(𝜓 → ∀𝑥𝜓) → Ⅎ𝑥𝜓)
53, 4syl 17 1 (𝜑 → Ⅎ𝑥𝜓)
 Colors of variables: wff setvar class Syntax hints:   → wi 4  ∀wal 1656  Ⅎwnf 1884 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1896  ax-4 1910  ax-10 2194 This theorem depends on definitions:  df-bi 199  df-ex 1881  df-nf 1885 This theorem is referenced by:  nf5dv  2201  hbimd  2332  ax12indalem  35021  ax12inda2ALT  35022
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