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Theorem bj-nnfand 37230
Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction, deduction form. Note: compared with the proof of bj-nnfan 37229, 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 37229 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 37230 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 1906 . . 3 (∃𝑥(𝜓𝜒) → (∃𝑥𝜓 ∧ ∃𝑥𝜒))
2 bj-nnfand.1 . . . . 5 (𝜑 → Ⅎ'𝑥𝜓)
32bj-nnfed 37207 . . . 4 (𝜑 → (∃𝑥𝜓𝜓))
4 bj-nnfand.2 . . . . 5 (𝜑 → Ⅎ'𝑥𝜒)
54bj-nnfed 37207 . . . 4 (𝜑 → (∃𝑥𝜒𝜒))
63, 5anim12d 618 . . 3 (𝜑 → ((∃𝑥𝜓 ∧ ∃𝑥𝜒) → (𝜓𝜒)))
71, 6syl5 34 . 2 (𝜑 → (∃𝑥(𝜓𝜒) → (𝜓𝜒)))
82bj-nnfad 37204 . . . 4 (𝜑 → (𝜓 → ∀𝑥𝜓))
94bj-nnfad 37204 . . . 4 (𝜑 → (𝜒 → ∀𝑥𝜒))
108, 9anim12d 618 . . 3 (𝜑 → ((𝜓𝜒) → (∀𝑥𝜓 ∧ ∀𝑥𝜒)))
11 19.26 1890 . . 3 (∀𝑥(𝜓𝜒) ↔ (∀𝑥𝜓 ∧ ∀𝑥𝜒))
1210, 11imbitrrdi 254 . 2 (𝜑 → ((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
13 df-bj-nnf 37202 . 2 (Ⅎ'𝑥(𝜓𝜒) ↔ ((∃𝑥(𝜓𝜒) → (𝜓𝜒)) ∧ ((𝜓𝜒) → ∀𝑥(𝜓𝜒))))
147, 12, 13sylanbrc 592 1 (𝜑 → Ⅎ'𝑥(𝜓𝜒))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 399  wal 1558  wex 1799  Ⅎ'wnnf 37201
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829
This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1800  df-bj-nnf 37202
This theorem is referenced by:  bj-nnfbid  37234
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