Metamath Proof Explorer


Theorem nf5d

Description: Deduce that x is not free in ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016)

Ref Expression
Hypotheses nf5d.1 ⊢ Ⅎ x φ
nf5d.2 ⊢ φ → ψ → ∀ x ψ
Assertion nf5d ⊢ φ → Ⅎ x ψ

Proof

Step Hyp Ref Expression
1 nf5d.1 ⊢ Ⅎ x φ
2 nf5d.2 ⊢ φ → ψ → ∀ x ψ
3 1 2 alrimi ⊢ φ → ∀ x ψ → ∀ x ψ
4 nf5-1 ⊢ ∀ x ψ → ∀ x ψ → Ⅎ x ψ
5 3 4 syl ⊢ φ → Ⅎ x ψ