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Theorem nf5dh 2190
Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Mario Carneiro, 24-Sep-2016.) df-nf 1864 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 1908 . 2 (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓))
4 nf5-1 2188 . 2 (∀𝑥(𝜓 → ∀𝑥𝜓) → Ⅎ𝑥𝜓)
53, 4syl 17 1 (𝜑 → Ⅎ𝑥𝜓)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wal 1635  wnf 1863
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-10 2184
This theorem depends on definitions:  df-bi 198  df-ex 1860  df-nf 1864
This theorem is referenced by:  nf5dv  2191  hbimd  2301  ax12indalem  34718  ax12inda2ALT  34719
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