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Theorem nfand 1528
Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, it is not free in (𝜓𝜒). (Contributed by Mario Carneiro, 7-Oct-2016.)
Hypotheses
Ref Expression
nfand.1 (𝜑 → Ⅎ𝑥𝜓)
nfand.2 (𝜑 → Ⅎ𝑥𝜒)
Assertion
Ref Expression
nfand (𝜑 → Ⅎ𝑥(𝜓𝜒))

Proof of Theorem nfand
StepHypRef Expression
1 nfand.1 . . . 4 (𝜑 → Ⅎ𝑥𝜓)
2 nfand.2 . . . 4 (𝜑 → Ⅎ𝑥𝜒)
31, 2jca 302 . . 3 (𝜑 → (Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒))
4 df-nf 1418 . . . . . 6 (Ⅎ𝑥𝜓 ↔ ∀𝑥(𝜓 → ∀𝑥𝜓))
5 df-nf 1418 . . . . . 6 (Ⅎ𝑥𝜒 ↔ ∀𝑥(𝜒 → ∀𝑥𝜒))
64, 5anbi12i 453 . . . . 5 ((Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒) ↔ (∀𝑥(𝜓 → ∀𝑥𝜓) ∧ ∀𝑥(𝜒 → ∀𝑥𝜒)))
7 19.26 1438 . . . . 5 (∀𝑥((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) ↔ (∀𝑥(𝜓 → ∀𝑥𝜓) ∧ ∀𝑥(𝜒 → ∀𝑥𝜒)))
86, 7bitr4i 186 . . . 4 ((Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒) ↔ ∀𝑥((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)))
9 prth 339 . . . . . 6 (((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) → ((𝜓𝜒) → (∀𝑥𝜓 ∧ ∀𝑥𝜒)))
10 19.26 1438 . . . . . 6 (∀𝑥(𝜓𝜒) ↔ (∀𝑥𝜓 ∧ ∀𝑥𝜒))
119, 10syl6ibr 161 . . . . 5 (((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) → ((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
1211alimi 1412 . . . 4 (∀𝑥((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
138, 12sylbi 120 . . 3 ((Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒) → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
143, 13syl 14 . 2 (𝜑 → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
15 df-nf 1418 . 2 (Ⅎ𝑥(𝜓𝜒) ↔ ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
1614, 15sylibr 133 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103  wal 1310  wnf 1417
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1404  ax-gen 1406
This theorem depends on definitions:  df-bi 116  df-nf 1418
This theorem is referenced by:  nf3and  1529  nfbid  1548  nfsbxy  1891  nfsbxyt  1892  nfeld  2269  nfrexdxy  2440  nfreudxy  2576  nfifd  3463  nfriotadxy  5690  bdsepnft  12768
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