Theorem bj-nnfand 36135

Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction, deduction form. Note: compared with the proof of bj-nnfan 36134, 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 36134 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 36135 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(𝜓 ∧ 𝜒))

Proof of Theorem bj-nnfand

Step	Hyp	Ref	Expression
1		19.40 1881	. . 3 ⊢ (∃𝑥(𝜓 ∧ 𝜒) → (∃𝑥𝜓 ∧ ∃𝑥𝜒))
2		bj-nnfand.1	. . . . 5 ⊢ (𝜑 → Ⅎ'𝑥𝜓)
3	2	bj-nnfed 36118	. . . 4 ⊢ (𝜑 → (∃𝑥𝜓 → 𝜓))
4		bj-nnfand.2	. . . . 5 ⊢ (𝜑 → Ⅎ'𝑥𝜒)
5	4	bj-nnfed 36118	. . . 4 ⊢ (𝜑 → (∃𝑥𝜒 → 𝜒))
6	3, 5	anim12d 608	. . 3 ⊢ (𝜑 → ((∃𝑥𝜓 ∧ ∃𝑥𝜒) → (𝜓 ∧ 𝜒)))
7	1, 6	syl5 34	. 2 ⊢ (𝜑 → (∃𝑥(𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜒)))
8	2	bj-nnfad 36115	. . . 4 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))
9	4	bj-nnfad 36115	. . . 4 ⊢ (𝜑 → (𝜒 → ∀𝑥𝜒))
10	8, 9	anim12d 608	. . 3 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) → (∀𝑥𝜓 ∧ ∀𝑥𝜒)))
11		19.26 1865	. . 3 ⊢ (∀𝑥(𝜓 ∧ 𝜒) ↔ (∀𝑥𝜓 ∧ ∀𝑥𝜒))
12	10, 11	imbitrrdi 251	. 2 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) → ∀𝑥(𝜓 ∧ 𝜒)))
13		df-bj-nnf 36110	. 2 ⊢ (Ⅎ'𝑥(𝜓 ∧ 𝜒) ↔ ((∃𝑥(𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜒)) ∧ ((𝜓 ∧ 𝜒) → ∀𝑥(𝜓 ∧ 𝜒))))
14	7, 12, 13	sylanbrc 582	1 ⊢ (𝜑 → Ⅎ'𝑥(𝜓 ∧ 𝜒))

Syntax hints:   → wi 4   ∧ wa 395  ∀wal 1531  ∃wex 1773  Ⅎ'wnnf 36109

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-bj-nnf 36110

This theorem is referenced by:  bj-nnfbid  36139