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Theorem bj-nnfand 34976
Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction, deduction form. Note: compared with the proof of bj-nnfan 34975, 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 34975 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 34976 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 1887 . . 3 (∃𝑥(𝜓𝜒) → (∃𝑥𝜓 ∧ ∃𝑥𝜒))
2 bj-nnfand.1 . . . . 5 (𝜑 → Ⅎ'𝑥𝜓)
32bj-nnfed 34959 . . . 4 (𝜑 → (∃𝑥𝜓𝜓))
4 bj-nnfand.2 . . . . 5 (𝜑 → Ⅎ'𝑥𝜒)
54bj-nnfed 34959 . . . 4 (𝜑 → (∃𝑥𝜒𝜒))
63, 5anim12d 610 . . 3 (𝜑 → ((∃𝑥𝜓 ∧ ∃𝑥𝜒) → (𝜓𝜒)))
71, 6syl5 34 . 2 (𝜑 → (∃𝑥(𝜓𝜒) → (𝜓𝜒)))
82bj-nnfad 34956 . . . 4 (𝜑 → (𝜓 → ∀𝑥𝜓))
94bj-nnfad 34956 . . . 4 (𝜑 → (𝜒 → ∀𝑥𝜒))
108, 9anim12d 610 . . 3 (𝜑 → ((𝜓𝜒) → (∀𝑥𝜓 ∧ ∀𝑥𝜒)))
11 19.26 1871 . . 3 (∀𝑥(𝜓𝜒) ↔ (∀𝑥𝜓 ∧ ∀𝑥𝜒))
1210, 11syl6ibr 252 . 2 (𝜑 → ((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
13 df-bj-nnf 34951 . 2 (Ⅎ'𝑥(𝜓𝜒) ↔ ((∃𝑥(𝜓𝜒) → (𝜓𝜒)) ∧ ((𝜓𝜒) → ∀𝑥(𝜓𝜒))))
147, 12, 13sylanbrc 584 1 (𝜑 → Ⅎ'𝑥(𝜓𝜒))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 397  wal 1537  wex 1779  Ⅎ'wnnf 34950
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809
This theorem depends on definitions:  df-bi 206  df-an 398  df-ex 1780  df-bj-nnf 34951
This theorem is referenced by:  bj-nnfbid  34980
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