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Theorem nford 1456
Description: If in a context x is not free in ψ and χ, it is not free in (ψ χ). (Contributed by Jim Kingdon, 29-Oct-2019.)
Hypotheses
Ref Expression
nford.1 (φ → Ⅎxψ)
nford.2 (φ → Ⅎxχ)
Assertion
Ref Expression
nford (φ → Ⅎx(ψ χ))

Proof of Theorem nford
StepHypRef Expression
1 nford.1 . . . . 5 (φ → Ⅎxψ)
2 nford.2 . . . . 5 (φ → Ⅎxχ)
3 df-nf 1347 . . . . . . 7 (Ⅎxψx(ψxψ))
4 df-nf 1347 . . . . . . 7 (Ⅎxχx(χxχ))
53, 4anbi12i 433 . . . . . 6 ((Ⅎxψ xχ) ↔ (x(ψxψ) x(χxχ)))
65biimpi 113 . . . . 5 ((Ⅎxψ xχ) → (x(ψxψ) x(χxχ)))
71, 2, 6syl2anc 391 . . . 4 (φ → (x(ψxψ) x(χxχ)))
8 19.26 1367 . . . 4 (x((ψxψ) (χxχ)) ↔ (x(ψxψ) x(χxχ)))
97, 8sylibr 137 . . 3 (φx((ψxψ) (χxχ)))
10 orc 632 . . . . . . 7 (ψ → (ψ χ))
1110alimi 1341 . . . . . 6 (xψx(ψ χ))
1211imim2i 12 . . . . 5 ((ψxψ) → (ψx(ψ χ)))
13 olc 631 . . . . . . 7 (χ → (ψ χ))
1413alimi 1341 . . . . . 6 (xχx(ψ χ))
1514imim2i 12 . . . . 5 ((χxχ) → (χx(ψ χ)))
1612, 15jaao 638 . . . 4 (((ψxψ) (χxχ)) → ((ψ χ) → x(ψ χ)))
1716alimi 1341 . . 3 (x((ψxψ) (χxχ)) → x((ψ χ) → x(ψ χ)))
189, 17syl 14 . 2 (φx((ψ χ) → x(ψ χ)))
19 df-nf 1347 . 2 (Ⅎx(ψ χ) ↔ x((ψ χ) → x(ψ χ)))
2018, 19sylibr 137 1 (φ → Ⅎx(ψ χ))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97   wo 628  wal 1240  wnf 1346
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-io 629  ax-5 1333  ax-gen 1335
This theorem depends on definitions:  df-bi 110  df-nf 1347
This theorem is referenced by:  nfifd  3349
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