Metamath Proof Explorer


Theorem nfexd

Description: If x is not free in ps , then it is not free in E. y ps . (Contributed by Mario Carneiro, 24-Sep-2016)

Ref Expression
Hypotheses nfald.1 ⊢ Ⅎ y φ
nfald.2 ⊢ φ → Ⅎ x ψ
Assertion nfexd ⊢ φ → Ⅎ x ∃ y ψ

Proof

Step Hyp Ref Expression
1 nfald.1 ⊢ Ⅎ y φ
2 nfald.2 ⊢ φ → Ⅎ x ψ
3 df-ex ⊢ ∃ y ψ ↔ ¬ ∀ y ¬ ψ
4 2 nfnd ⊢ φ → Ⅎ x ¬ ψ
5 1 4 nfald ⊢ φ → Ⅎ x ∀ y ¬ ψ
6 5 nfnd ⊢ φ → Ⅎ x ¬ ∀ y ¬ ψ
7 3 6 nfxfrd ⊢ φ → Ⅎ x ∃ y ψ