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

Step	Hyp	Ref	Expression
1		nfd2.1	. 2 ⊢ (𝜑 → (∃𝑥𝜓 → ∀𝑥𝜓))
2		nf2 1668	. 2 ⊢ (Ⅎ𝑥𝜓 ↔ (∃𝑥𝜓 → ∀𝑥𝜓))
3	1, 2	sylibr 134	1 ⊢ (𝜑 → Ⅎ𝑥𝜓)

Syntax hints:   → wi 4  ∀wal 1351  Ⅎwnf 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