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Theorem nfd2 2022
Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Wolf Lammen, 16-Sep-2021.)
Hypothesis
Ref Expression
nfd2.1 (𝜑 → (∃𝑥𝜓 → ∀𝑥𝜓))
Assertion
Ref Expression
nfd2 (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nfd2
StepHypRef Expression
1 nfd2.1 . 2 (𝜑 → (∃𝑥𝜓 → ∀𝑥𝜓))
2 nf2 1668 . 2 (Ⅎ𝑥𝜓 ↔ (∃𝑥𝜓 → ∀𝑥𝜓))
31, 2sylibr 134 1 (𝜑 → Ⅎ𝑥𝜓)
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1351  wnf 1460  wex 1492
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-gen 1449  ax-ie2 1494  ax-4 1510  ax-ial 1534
This theorem depends on definitions:  df-bi 117  df-nf 1461
This theorem is referenced by:  nf5-1  2024
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