Metamath Proof Explorer


Theorem nfd

Description: Deduce that x is not free in ps in a context. (Contributed by Wolf Lammen, 16-Sep-2021)

Ref Expression
Hypothesis nfd.1 ⊢ φ → ∃ x ψ → ∀ x ψ
Assertion nfd ⊢ φ → Ⅎ x ψ

Proof

Step Hyp Ref Expression
1 nfd.1 ⊢ φ → ∃ x ψ → ∀ x ψ
2 df-nf ⊢ Ⅎ x ψ ↔ ∃ x ψ → ∀ x ψ
3 1 2 sylibr ⊢ φ → Ⅎ x ψ