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	$⊢ φ \to Ⅎ x ψ$
	nfimd.2	$⊢ φ \to Ⅎ x χ$
Assertion	nfimd	$⊢ φ \to Ⅎ x (ψ \to χ)$

Proof

Step	Hyp	Ref	Expression
1		nfimd.1	$⊢ φ \to Ⅎ x ψ$
2		nfimd.2	$⊢ φ \to Ⅎ x χ$
3		19.35	$⊢ \exists x (ψ \to χ) \leftrightarrow (\forall x ψ \to \exists x χ)$
4	3	biimpi	$⊢ \exists x (ψ \to χ) \to (\forall x ψ \to \exists x χ)$
5	1	nfrd	$⊢ φ \to (\exists x ψ \to \forall x ψ)$
6	2	nfrd	$⊢ φ \to (\exists x χ \to \forall x χ)$
7	5 6	imim12d	$⊢ φ \to ((\forall x ψ \to \exists x χ) \to (\exists x ψ \to \forall x χ))$
8		19.38	$⊢ (\exists x ψ \to \forall x χ) \to \forall x (ψ \to χ)$
9	4 7 8	syl56	$⊢ φ \to (\exists x (ψ \to χ) \to \forall x (ψ \to χ))$
10	9	nfd	$⊢ φ \to Ⅎ x (ψ \to χ)$

Metamath Proof Explorer

Theorem nfimd

Proof