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Theorem bj-nnfand 37268
Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction, deduction form. Note: compared with the proof of bj-nnfan 37267, 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 37267 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 37268 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 1913 . . 3 (∃𝑥(𝜓𝜒) → (∃𝑥𝜓 ∧ ∃𝑥𝜒))
2 bj-nnfand.1 . . . . 5 (𝜑 → Ⅎ'𝑥𝜓)
32bj-nnfed 37245 . . . 4 (𝜑 → (∃𝑥𝜓𝜓))
4 bj-nnfand.2 . . . . 5 (𝜑 → Ⅎ'𝑥𝜒)
54bj-nnfed 37245 . . . 4 (𝜑 → (∃𝑥𝜒𝜒))
63, 5anim12d 620 . . 3 (𝜑 → ((∃𝑥𝜓 ∧ ∃𝑥𝜒) → (𝜓𝜒)))
71, 6syl5 35 . 2 (𝜑 → (∃𝑥(𝜓𝜒) → (𝜓𝜒)))
82bj-nnfad 37242 . . . 4 (𝜑 → (𝜓 → ∀𝑥𝜓))
94bj-nnfad 37242 . . . 4 (𝜑 → (𝜒 → ∀𝑥𝜒))
108, 9anim12d 620 . . 3 (𝜑 → ((𝜓𝜒) → (∀𝑥𝜓 ∧ ∀𝑥𝜒)))
11 19.26 1897 . . 3 (∀𝑥(𝜓𝜒) ↔ (∀𝑥𝜓 ∧ ∀𝑥𝜒))
1210, 11imbitrrdi 255 . 2 (𝜑 → ((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
13 df-bj-nnf 37240 . 2 (Ⅎ'𝑥(𝜓𝜒) ↔ ((∃𝑥(𝜓𝜒) → (𝜓𝜒)) ∧ ((𝜓𝜒) → ∀𝑥(𝜓𝜒))))
147, 12, 13sylanbrc 594 1 (𝜑 → Ⅎ'𝑥(𝜓𝜒))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 400  wal 1565  wex 1806  Ⅎ'wnnf 37239
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836
This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1807  df-bj-nnf 37240
This theorem is referenced by:  bj-nnfbid  37272
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