Theorem bj-nnfand 36789

Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction, deduction form. Note: compared with the proof of bj-nnfan 36788, 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 36788 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 36789 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 1887	. . 3 ⊢ (∃𝑥(𝜓 ∧ 𝜒) → (∃𝑥𝜓 ∧ ∃𝑥𝜒))
2		bj-nnfand.1	. . . . 5 ⊢ (𝜑 → Ⅎ'𝑥𝜓)
3	2	bj-nnfed 36772	. . . 4 ⊢ (𝜑 → (∃𝑥𝜓 → 𝜓))
4		bj-nnfand.2	. . . . 5 ⊢ (𝜑 → Ⅎ'𝑥𝜒)
5	4	bj-nnfed 36772	. . . 4 ⊢ (𝜑 → (∃𝑥𝜒 → 𝜒))
6	3, 5	anim12d 609	. . 3 ⊢ (𝜑 → ((∃𝑥𝜓 ∧ ∃𝑥𝜒) → (𝜓 ∧ 𝜒)))
7	1, 6	syl5 34	. 2 ⊢ (𝜑 → (∃𝑥(𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜒)))
8	2	bj-nnfad 36769	. . . 4 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))
9	4	bj-nnfad 36769	. . . 4 ⊢ (𝜑 → (𝜒 → ∀𝑥𝜒))
10	8, 9	anim12d 609	. . 3 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) → (∀𝑥𝜓 ∧ ∀𝑥𝜒)))
11		19.26 1871	. . 3 ⊢ (∀𝑥(𝜓 ∧ 𝜒) ↔ (∀𝑥𝜓 ∧ ∀𝑥𝜒))
12	10, 11	imbitrrdi 252	. 2 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) → ∀𝑥(𝜓 ∧ 𝜒)))
13		df-bj-nnf 36764	. 2 ⊢ (Ⅎ'𝑥(𝜓 ∧ 𝜒) ↔ ((∃𝑥(𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜒)) ∧ ((𝜓 ∧ 𝜒) → ∀𝑥(𝜓 ∧ 𝜒))))
14	7, 12, 13	sylanbrc 583	1 ⊢ (𝜑 → Ⅎ'𝑥(𝜓 ∧ 𝜒))

Syntax hints:   → wi 4   ∧ wa 395  ∀wal 1539  ∃wex 1780  Ⅎ'wnnf 36763

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1781  df-bj-nnf 36764

This theorem is referenced by:  bj-nnfbid  36793