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Theorem nfand 1438
Description: If in a context x is not free in ψ and χ, it is not free in (ψ χ). (Contributed by Mario Carneiro, 7-Oct-2016.)
Hypotheses
Ref Expression
nfand.1 (φ → Ⅎxψ)
nfand.2 (φ → Ⅎxχ)
Assertion
Ref Expression
nfand (φ → Ⅎx(ψ χ))

Proof of Theorem nfand
StepHypRef Expression
1 nfand.1 . . . 4 (φ → Ⅎxψ)
2 nfand.2 . . . 4 (φ → Ⅎxχ)
31, 2jca 290 . . 3 (φ → (Ⅎxψ xχ))
4 df-nf 1326 . . . . . 6 (Ⅎxψx(ψxψ))
5 df-nf 1326 . . . . . 6 (Ⅎxχx(χxχ))
64, 5anbi12i 436 . . . . 5 ((Ⅎxψ xχ) ↔ (x(ψxψ) x(χxχ)))
7 19.26 1346 . . . . 5 (x((ψxψ) (χxχ)) ↔ (x(ψxψ) x(χxχ)))
86, 7bitr4i 176 . . . 4 ((Ⅎxψ xχ) ↔ x((ψxψ) (χxχ)))
9 prth 326 . . . . . 6 (((ψxψ) (χxχ)) → ((ψ χ) → (xψ xχ)))
10 19.26 1346 . . . . . 6 (x(ψ χ) ↔ (xψ xχ))
119, 10syl6ibr 151 . . . . 5 (((ψxψ) (χxχ)) → ((ψ χ) → x(ψ χ)))
1211alimi 1320 . . . 4 (x((ψxψ) (χxχ)) → x((ψ χ) → x(ψ χ)))
138, 12sylbi 114 . . 3 ((Ⅎxψ xχ) → x((ψ χ) → x(ψ χ)))
143, 13syl 14 . 2 (φx((ψ χ) → x(ψ χ)))
15 df-nf 1326 . 2 (Ⅎx(ψ χ) ↔ x((ψ χ) → x(ψ χ)))
1614, 15sylibr 137 1 (φ → Ⅎx(ψ χ))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wal 1224  wnf 1325
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 1312  ax-gen 1314
This theorem depends on definitions:  df-bi 110  df-nf 1326
This theorem is referenced by:  nf3and  1439  nfbid  1458  nfsbxy  1796  nfsbxyt  1797  nfeld  2171  nfrexdxy  2331  nfreudxy  2457  nfifd  3328  nfriotadxy  5396  bdsepnft  7252
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