Theorem nfimd 1596

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 1555	. . . . 5 ⊢ Ⅎ𝑥Ⅎ𝑥𝜓
4	3	nfri 1530	. . . 4 ⊢ (Ⅎ𝑥𝜓 → ∀𝑥Ⅎ𝑥𝜓)
5		nfnf1 1555	. . . . 5 ⊢ Ⅎ𝑥Ⅎ𝑥𝜒
6	5	nfri 1530	. . . 4 ⊢ (Ⅎ𝑥𝜒 → ∀𝑥Ⅎ𝑥𝜒)
7		nfr 1529	. . . . . 6 ⊢ (Ⅎ𝑥𝜒 → (𝜒 → ∀𝑥𝜒))
8	7	imim2d 54	. . . . 5 ⊢ (Ⅎ𝑥𝜒 → ((𝜓 → 𝜒) → (𝜓 → ∀𝑥𝜒)))
9		19.21t 1593	. . . . . 6 ⊢ (Ⅎ𝑥𝜓 → (∀𝑥(𝜓 → 𝜒) ↔ (𝜓 → ∀𝑥𝜒)))
10	9	biimprd 158	. . . . 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 1490	. . 3 ⊢ (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → ∀𝑥((𝜓 → 𝜒) → ∀𝑥(𝜓 → 𝜒))))
13		df-nf 1472	. . 3 ⊢ (Ⅎ𝑥(𝜓 → 𝜒) ↔ ∀𝑥((𝜓 → 𝜒) → ∀𝑥(𝜓 → 𝜒)))
14	12, 13	imbitrrdi 162	. 2 ⊢ (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → Ⅎ𝑥(𝜓 → 𝜒)))
15	1, 2, 14	sylc 62	1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → 𝜒))

Syntax hints:   → wi 4  ∀wal 1362  Ⅎwnf 1471

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1458  ax-gen 1460  ax-4 1521  ax-ial 1545  ax-i5r 1546

This theorem depends on definitions:  df-bi 117  df-nf 1472

This theorem is referenced by:  nfbid  1599  dvelimALT  2022  dvelimfv  2023  dvelimor  2030  nfmod  2055  nfraldw  2522  nfraldxy  2523  nfixpxy  6744  cbvrald  15018