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Theorem nfd2 2015
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 1661 . 2 (Ⅎ𝑥𝜓 ↔ (∃𝑥𝜓 → ∀𝑥𝜓))
31, 2sylibr 133 1 (𝜑 → Ⅎ𝑥𝜓)
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1346  wnf 1453  wex 1485
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-gen 1442  ax-ie2 1487  ax-4 1503  ax-ial 1527
This theorem depends on definitions:  df-bi 116  df-nf 1454
This theorem is referenced by:  nf5-1  2017
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