Metamath Proof Explorer


Theorem nfimd

Description: If in a context x is not free in ps and ch , then it is not free in ( ps -> ch ) . Deduction form of nfim . (Contributed by Mario Carneiro, 24-Sep-2016) (Proof shortened by Wolf Lammen, 30-Dec-2017) df-nf changed. (Revised by Wolf Lammen, 18-Sep-2021) Eliminate curried form of nfimt . (Revised by Wolf Lammen, 10-Jul-2022)

Ref Expression
Hypotheses nfimd.1 ⊢ φ → Ⅎ x ψ
nfimd.2 ⊢ φ → Ⅎ x χ
Assertion nfimd ⊢ φ → Ⅎ x ψ → χ

Proof

Step Hyp Ref Expression
1 nfimd.1 ⊢ φ → Ⅎ x ψ
2 nfimd.2 ⊢ φ → Ⅎ x χ
3 19.35 ⊢ ∃ x ψ → χ ↔ ∀ x ψ → ∃ x χ
4 3 biimpi ⊢ ∃ x ψ → χ → ∀ x ψ → ∃ x χ
5 1 nfrd ⊢ φ → ∃ x ψ → ∀ x ψ
6 2 nfrd ⊢ φ → ∃ x χ → ∀ x χ
7 5 6 imim12d ⊢ φ → ∀ x ψ → ∃ x χ → ∃ x ψ → ∀ x χ
8 19.38 ⊢ ∃ x ψ → ∀ x χ → ∀ x ψ → χ
9 4 7 8 syl56 ⊢ φ → ∃ x ψ → χ → ∀ x ψ → χ
10 9 nfd ⊢ φ → Ⅎ x ψ → χ