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Theorem bj-nnfand 37489
Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction, deduction form. Note: compared with the proof of bj-nnfan 37488, it has two more essential steps but fewer total steps (since there are fewer intermediate formulas to build) and is easier to follow and understand. This statement is of intermediate complexity: for simpler statements, closed-style proofs like that of bj-nnfan 37488 will generally be shorter than deduction-style proofs while still easy to follow, while for more complex statements, the opposite will be true (and deduction-style proofs like that of bj-nnfand 37489 will generally be easier to understand). (Contributed by BJ, 19-Nov-2023.) (Proof modification is discouraged.)
Hypotheses
Ref Expression
bj-nnfand.1 (𝜑 → Ⅎ'𝑥𝜓)
bj-nnfand.2 (𝜑 → Ⅎ'𝑥𝜒)
Assertion
Ref Expression
bj-nnfand (𝜑 → Ⅎ'𝑥(𝜓𝜒))

Proof of Theorem bj-nnfand
StepHypRef Expression
1 19.40 1919 . . 3 (∃𝑥(𝜓𝜒) → (∃𝑥𝜓 ∧ ∃𝑥𝜒))
2 bj-nnfand.1 . . . . 5 (𝜑 → Ⅎ'𝑥𝜓)
32bj-nnfed 37466 . . . 4 (𝜑 → (∃𝑥𝜓𝜓))
4 bj-nnfand.2 . . . . 5 (𝜑 → Ⅎ'𝑥𝜒)
54bj-nnfed 37466 . . . 4 (𝜑 → (∃𝑥𝜒𝜒))
63, 5anim12d 621 . . 3 (𝜑 → ((∃𝑥𝜓 ∧ ∃𝑥𝜒) → (𝜓𝜒)))
71, 6syl5 35 . 2 (𝜑 → (∃𝑥(𝜓𝜒) → (𝜓𝜒)))
82bj-nnfad 37463 . . . 4 (𝜑 → (𝜓 → ∀𝑥𝜓))
94bj-nnfad 37463 . . . 4 (𝜑 → (𝜒 → ∀𝑥𝜒))
108, 9anim12d 621 . . 3 (𝜑 → ((𝜓𝜒) → (∀𝑥𝜓 ∧ ∀𝑥𝜒)))
11 19.26 1903 . . 3 (∀𝑥(𝜓𝜒) ↔ (∀𝑥𝜓 ∧ ∀𝑥𝜒))
1210, 11imbitrrdi 255 . 2 (𝜑 → ((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
13 df-bj-nnf 37461 . 2 (Ⅎ'𝑥(𝜓𝜒) ↔ ((∃𝑥(𝜓𝜒) → (𝜓𝜒)) ∧ ((𝜓𝜒) → ∀𝑥(𝜓𝜒))))
147, 12, 13sylanbrc 595 1 (𝜑 → Ⅎ'𝑥(𝜓𝜒))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wa 401  wal 1568  wex 1812  Ⅎ'wnnf 37460
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842
This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-bj-nnf 37461
This theorem is used by:  bj-nnfbid  37493
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