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Theorem nfand 1501
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 300 . . 3 (𝜑 → (Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒))
4 df-nf 1391 . . . . . 6 (Ⅎ𝑥𝜓 ↔ ∀𝑥(𝜓 → ∀𝑥𝜓))
5 df-nf 1391 . . . . . 6 (Ⅎ𝑥𝜒 ↔ ∀𝑥(𝜒 → ∀𝑥𝜒))
64, 5anbi12i 448 . . . . 5 ((Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒) ↔ (∀𝑥(𝜓 → ∀𝑥𝜓) ∧ ∀𝑥(𝜒 → ∀𝑥𝜒)))
7 19.26 1411 . . . . 5 (∀𝑥((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) ↔ (∀𝑥(𝜓 → ∀𝑥𝜓) ∧ ∀𝑥(𝜒 → ∀𝑥𝜒)))
86, 7bitr4i 185 . . . 4 ((Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒) ↔ ∀𝑥((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)))
9 prth 336 . . . . . 6 (((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) → ((𝜓𝜒) → (∀𝑥𝜓 ∧ ∀𝑥𝜒)))
10 19.26 1411 . . . . . 6 (∀𝑥(𝜓𝜒) ↔ (∀𝑥𝜓 ∧ ∀𝑥𝜒))
119, 10syl6ibr 160 . . . . 5 (((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) → ((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
1211alimi 1385 . . . 4 (∀𝑥((𝜓 → ∀𝑥𝜓) ∧ (𝜒 → ∀𝑥𝜒)) → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
138, 12sylbi 119 . . 3 ((Ⅎ𝑥𝜓 ∧ Ⅎ𝑥𝜒) → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
143, 13syl 14 . 2 (𝜑 → ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
15 df-nf 1391 . 2 (Ⅎ𝑥(𝜓𝜒) ↔ ∀𝑥((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
1614, 15sylibr 132 1 (𝜑 → Ⅎ𝑥(𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wal 1283  wnf 1390
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1377  ax-gen 1379
This theorem depends on definitions:  df-bi 115  df-nf 1391
This theorem is referenced by:  nf3and  1502  nfbid  1521  nfsbxy  1860  nfsbxyt  1861  nfeld  2235  nfrexdxy  2400  nfreudxy  2528  nfifd  3384  nfriotadxy  5507  bdsepnft  10836
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