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Theorem nfimd 1564
 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(𝜓𝜒))

Proof of Theorem nfimd
StepHypRef Expression
1 nfimd.1 . 2 (𝜑 → Ⅎ𝑥𝜓)
2 nfimd.2 . 2 (𝜑 → Ⅎ𝑥𝜒)
3 nfnf1 1523 . . . . 5 𝑥𝑥𝜓
43nfri 1499 . . . 4 (Ⅎ𝑥𝜓 → ∀𝑥𝑥𝜓)
5 nfnf1 1523 . . . . 5 𝑥𝑥𝜒
65nfri 1499 . . . 4 (Ⅎ𝑥𝜒 → ∀𝑥𝑥𝜒)
7 nfr 1498 . . . . . 6 (Ⅎ𝑥𝜒 → (𝜒 → ∀𝑥𝜒))
87imim2d 54 . . . . 5 (Ⅎ𝑥𝜒 → ((𝜓𝜒) → (𝜓 → ∀𝑥𝜒)))
9 19.21t 1561 . . . . . 6 (Ⅎ𝑥𝜓 → (∀𝑥(𝜓𝜒) ↔ (𝜓 → ∀𝑥𝜒)))
109biimprd 157 . . . . 5 (Ⅎ𝑥𝜓 → ((𝜓 → ∀𝑥𝜒) → ∀𝑥(𝜓𝜒)))
118, 10syl9r 73 . . . 4 (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → ((𝜓𝜒) → ∀𝑥(𝜓𝜒))))
124, 6, 11alrimdh 1455 . . 3 (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒))))
13 df-nf 1437 . . 3 (Ⅎ𝑥(𝜓𝜒) ↔ ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
1412, 13syl6ibr 161 . 2 (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → Ⅎ𝑥(𝜓𝜒)))
151, 2, 14sylc 62 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
 Colors of variables: wff set class Syntax hints:   → wi 4  ∀wal 1329  Ⅎwnf 1436 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 1423  ax-gen 1425  ax-4 1487  ax-ial 1514  ax-i5r 1515 This theorem depends on definitions:  df-bi 116  df-nf 1437 This theorem is referenced by:  nfbid  1567  dvelimALT  1983  dvelimfv  1984  dvelimor  1991  nfmod  2014  nfraldxy  2465  nfixpxy  6604  cbvrald  12984
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