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Theorem nfimd 1522
Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, then it is not free in (𝜓𝜒). (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 1481 . . . . 5 𝑥𝑥𝜓
43nfri 1457 . . . 4 (Ⅎ𝑥𝜓 → ∀𝑥𝑥𝜓)
5 nfnf1 1481 . . . . 5 𝑥𝑥𝜒
65nfri 1457 . . . 4 (Ⅎ𝑥𝜒 → ∀𝑥𝑥𝜒)
7 nfr 1456 . . . . . 6 (Ⅎ𝑥𝜒 → (𝜒 → ∀𝑥𝜒))
87imim2d 53 . . . . 5 (Ⅎ𝑥𝜒 → ((𝜓𝜒) → (𝜓 → ∀𝑥𝜒)))
9 19.21t 1519 . . . . . 6 (Ⅎ𝑥𝜓 → (∀𝑥(𝜓𝜒) ↔ (𝜓 → ∀𝑥𝜒)))
109biimprd 156 . . . . 5 (Ⅎ𝑥𝜓 → ((𝜓 → ∀𝑥𝜒) → ∀𝑥(𝜓𝜒)))
118, 10syl9r 72 . . . 4 (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → ((𝜓𝜒) → ∀𝑥(𝜓𝜒))))
124, 6, 11alrimdh 1413 . . 3 (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒))))
13 df-nf 1395 . . 3 (Ⅎ𝑥(𝜓𝜒) ↔ ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
1412, 13syl6ibr 160 . 2 (Ⅎ𝑥𝜓 → (Ⅎ𝑥𝜒 → Ⅎ𝑥(𝜓𝜒)))
151, 2, 14sylc 61 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1287  wnf 1394
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1381  ax-gen 1383  ax-4 1445  ax-ial 1472  ax-i5r 1473
This theorem depends on definitions:  df-bi 115  df-nf 1395
This theorem is referenced by:  nfbid  1525  dvelimALT  1934  dvelimfv  1935  dvelimor  1942  nfmod  1965  nfraldxy  2410  cbvrald  11643
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