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Theorem nfand 1579
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 306 . . 3 (𝜑 → (Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒))
4 df-nf 1472 . . . . . 6 (Ⅎ𝑥𝜓 ↔ ∀𝑥(𝜓 → ∀𝑥𝜓))
5 df-nf 1472 . . . . . 6 (Ⅎ𝑥𝜒 ↔ ∀𝑥(𝜒 → ∀𝑥𝜒))
64, 5anbi12i 460 . . . . 5 ((Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒) ↔ (∀𝑥(𝜓 → ∀𝑥𝜓) ∧ ∀𝑥(𝜒 → ∀𝑥𝜒)))
7 19.26 1492 . . . . 5 (∀𝑥((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) ↔ (∀𝑥(𝜓 → ∀𝑥𝜓) ∧ ∀𝑥(𝜒 → ∀𝑥𝜒)))
86, 7bitr4i 187 . . . 4 ((Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒) ↔ ∀𝑥((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)))
9 anim12 344 . . . . . 6 (((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) → ((𝜓𝜒) → (∀𝑥𝜓 ∧ ∀𝑥𝜒)))
10 19.26 1492 . . . . . 6 (∀𝑥(𝜓𝜒) ↔ (∀𝑥𝜓 ∧ ∀𝑥𝜒))
119, 10imbitrrdi 162 . . . . 5 (((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) → ((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
1211alimi 1466 . . . 4 (∀𝑥((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
138, 12sylbi 121 . . 3 ((Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒) → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
143, 13syl 14 . 2 (𝜑 → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
15 df-nf 1472 . 2 (Ⅎ𝑥(𝜓𝜒) ↔ ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
1614, 15sylibr 134 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104  wal 1362  wnf 1471
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1458  ax-gen 1460
This theorem depends on definitions:  df-bi 117  df-nf 1472
This theorem is referenced by:  nf3and  1580  nfbid  1599  nfsbxy  1954  nfsbxyt  1955  nfeld  2348  nfrexdxy  2524  nfreudxy  2664  nfifd  3576  nfriotadxy  5861  bdsepnft  15117  strcollnft  15214
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