Theorem nfimd 1565

Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, then 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 1524	. . . . 5 ⊢ Ⅎ𝑥Ⅎ𝑥𝜓
4	3	nfri 1500	. . . 4 ⊢ (Ⅎ𝑥𝜓 → ∀𝑥Ⅎ𝑥𝜓)
5		nfnf1 1524	. . . . 5 ⊢ Ⅎ𝑥Ⅎ𝑥𝜒
6	5	nfri 1500	. . . 4 ⊢ (Ⅎ𝑥𝜒 → ∀𝑥Ⅎ𝑥𝜒)
7		nfr 1499	. . . . . 6 ⊢ (Ⅎ𝑥𝜒 → (𝜒 → ∀𝑥𝜒))
8	7	imim2d 54	. . . . 5 ⊢ (Ⅎ𝑥𝜒 → ((𝜓 → 𝜒) → (𝜓 → ∀𝑥𝜒)))
9		19.21t 1562	. . . . . 6 ⊢ (Ⅎ𝑥𝜓 → (∀𝑥(𝜓 → 𝜒) ↔ (𝜓 → ∀𝑥𝜒)))
10	9	biimprd 157	. . . . 5 ⊢ (Ⅎ𝑥𝜓 → ((𝜓 → ∀𝑥𝜒) → ∀𝑥(𝜓 → 𝜒)))
11	8, 10	syl9r 73	. . . 4 ⊢ (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → ((𝜓 → 𝜒) → ∀𝑥(𝜓 → 𝜒))))
12	4, 6, 11	alrimdh 1456	. . 3 ⊢ (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → ∀𝑥((𝜓 → 𝜒) → ∀𝑥(𝜓 → 𝜒))))
13		df-nf 1438	. . 3 ⊢ (Ⅎ𝑥(𝜓 → 𝜒) ↔ ∀𝑥((𝜓 → 𝜒) → ∀𝑥(𝜓 → 𝜒)))
14	12, 13	syl6ibr 161	. 2 ⊢ (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → Ⅎ𝑥(𝜓 → 𝜒)))
15	1, 2, 14	sylc 62	1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → 𝜒))

Syntax hints:   → wi 4  ∀wal 1330  Ⅎwnf 1437

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1424  ax-gen 1426  ax-4 1488  ax-ial 1515  ax-i5r 1516

This theorem depends on definitions:  df-bi 116  df-nf 1438

This theorem is referenced by:  nfbid  1568  dvelimALT  1986  dvelimfv  1987  dvelimor  1994  nfmod  2017  nfraldxy  2470  nfixpxy  6619  cbvrald  13166