Theorem nfimd 1808

Description: If in a context x is not free in ψ and χ, it is not free in (ψ → χ). (Contributed by Mario Carneiro, 24-Sep-2016.) (Proof shortened by Wolf Lammen, 30-Dec-2017.)

Hypotheses

Ref	Expression
nfimd.1	⊢ (φ → Ⅎxψ)
nfimd.2	⊢ (φ → Ⅎxχ)

Assertion

Ref	Expression
nfimd	⊢ (φ → Ⅎx(ψ → χ))

Proof of Theorem nfimd

Step	Hyp	Ref	Expression
1		nfimd.1	. 2 ⊢ (φ → Ⅎxψ)
2		nfimd.2	. 2 ⊢ (φ → Ⅎxχ)
3		nfnf1 1790	. . . 4 ⊢ ℲxℲxψ
4		nfnf1 1790	. . . 4 ⊢ ℲxℲxχ
5		nfr 1761	. . . . . 6 ⊢ (Ⅎxχ → (χ → ∀xχ))
6	5	imim2d 48	. . . . 5 ⊢ (Ⅎxχ → ((ψ → χ) → (ψ → ∀xχ)))
7		19.21t 1795	. . . . . 6 ⊢ (Ⅎxψ → (∀x(ψ → χ) ↔ (ψ → ∀xχ)))
8	7	biimprd 214	. . . . 5 ⊢ (Ⅎxψ → ((ψ → ∀xχ) → ∀x(ψ → χ)))
9	6, 8	syl9r 67	. . . 4 ⊢ (Ⅎxψ → (Ⅎxχ → ((ψ → χ) → ∀x(ψ → χ))))
10	3, 4, 9	alrimd 1769	. . 3 ⊢ (Ⅎxψ → (Ⅎxχ → ∀x((ψ → χ) → ∀x(ψ → χ))))
11		df-nf 1545	. . 3 ⊢ (Ⅎx(ψ → χ) ↔ ∀x((ψ → χ) → ∀x(ψ → χ)))
12	10, 11	syl6ibr 218	. 2 ⊢ (Ⅎxψ → (Ⅎxχ → Ⅎx(ψ → χ)))
13	1, 2, 12	sylc 56	1 ⊢ (φ → Ⅎx(ψ → χ))

This proof depends on syntax axioms:   → wi 4  ∀wal 1540  Ⅎwnf 1544

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-11 1746

This proof depends on definitions:  df-bi 177  df-ex 1542  df-nf 1545

This theorem is used by:  nfimOLD  1814  hbimd  1815  19.23tOLD  1819  nfand  1822  nfbid  1832  nfbidOLD  1833  nfnfOLD  1846  19.21tOLD  1863  dvelimf  1997  nfsb4t  2080  nfmod2  2217  nfrald  2666  nfifd  3686