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Theorem nfimd 1450
Description: If in a context x is not free in ψ and χ, 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 1409 . . . . 5 xxψ
43nfri 1385 . . . 4 (Ⅎxψxxψ)
5 nfnf1 1409 . . . . 5 xxχ
65nfri 1385 . . . 4 (Ⅎxχxxχ)
7 nfr 1384 . . . . . 6 (Ⅎxχ → (χxχ))
87imim2d 48 . . . . 5 (Ⅎxχ → ((ψχ) → (ψxχ)))
9 19.21t 1447 . . . . . 6 (Ⅎxψ → (x(ψχ) ↔ (ψxχ)))
109biimprd 147 . . . . 5 (Ⅎxψ → ((ψxχ) → x(ψχ)))
118, 10syl9r 67 . . . 4 (Ⅎxψ → (Ⅎxχ → ((ψχ) → x(ψχ))))
124, 6, 11alrimdh 1341 . . 3 (Ⅎxψ → (Ⅎxχx((ψχ) → x(ψχ))))
13 df-nf 1323 . . 3 (Ⅎx(ψχ) ↔ x((ψχ) → x(ψχ)))
1412, 13syl6ibr 151 . 2 (Ⅎxψ → (Ⅎxχ → Ⅎx(ψχ)))
151, 2, 14sylc 56 1 (φ → Ⅎx(ψχ))
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1221  wnf 1322
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1309  ax-gen 1311  ax-4 1373  ax-ial 1400  ax-i5r 1401
This theorem depends on definitions:  df-bi 110  df-nf 1323
This theorem is referenced by:  nfbid  1453  dvelimALT  1859  dvelimfv  1860  dvelimor  1867  nfmod  1890  nfraldxy  2325  cbvrald  8192
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