Theorem bj-nnfand 36772

Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction, deduction form. Note: compared with the proof of bj-nnfan 36771, 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 36771 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 36772 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 1886	. . 3 ⊢ (∃𝑥(𝜓 ∧ 𝜒) → (∃𝑥𝜓 ∧ ∃𝑥𝜒))
2		bj-nnfand.1	. . . . 5 ⊢ (𝜑 → Ⅎ'𝑥𝜓)
3	2	bj-nnfed 36755	. . . 4 ⊢ (𝜑 → (∃𝑥𝜓 → 𝜓))
4		bj-nnfand.2	. . . . 5 ⊢ (𝜑 → Ⅎ'𝑥𝜒)
5	4	bj-nnfed 36755	. . . 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 36752	. . . 4 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))
9	4	bj-nnfad 36752	. . . 4 ⊢ (𝜑 → (𝜒 → ∀𝑥𝜒))
10	8, 9	anim12d 609	. . 3 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) → (∀𝑥𝜓 ∧ ∀𝑥𝜒)))
11		19.26 1870	. . 3 ⊢ (∀𝑥(𝜓 ∧ 𝜒) ↔ (∀𝑥𝜓 ∧ ∀𝑥𝜒))
12	10, 11	imbitrrdi 252	. 2 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) → ∀𝑥(𝜓 ∧ 𝜒)))
13		df-bj-nnf 36747	. 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 1538  ∃wex 1779  Ⅎ'wnnf 36746

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 207  df-an 396  df-ex 1780  df-bj-nnf 36747

This theorem is referenced by:  bj-nnfbid  36776