Theorem bj-nnfan 36258

Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction. (Contributed by BJ, 19-Nov-2023.) In classical logic, there is a proof using the definition of conjunction in terms of implication and negation, so using bj-nnfim 36256, bj-nnfnt 36250 and bj-nnfbi 36235, but we want a proof valid in intuitionistic logic. (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-nnfan	⊢ ((Ⅎ'𝑥𝜑 ∧ Ⅎ'𝑥𝜓) → Ⅎ'𝑥(𝜑 ∧ 𝜓))

Proof of Theorem bj-nnfan

Step	Hyp	Ref	Expression
1		df-bj-nnf 36234	. . 3 ⊢ (Ⅎ'𝑥𝜑 ↔ ((∃𝑥𝜑 → 𝜑) ∧ (𝜑 → ∀𝑥𝜑)))
2		df-bj-nnf 36234	. . 3 ⊢ (Ⅎ'𝑥𝜓 ↔ ((∃𝑥𝜓 → 𝜓) ∧ (𝜓 → ∀𝑥𝜓)))
3		19.40 1881	. . . . . 6 ⊢ (∃𝑥(𝜑 ∧ 𝜓) → (∃𝑥𝜑 ∧ ∃𝑥𝜓))
4		anim12 807	. . . . . 6 ⊢ (((∃𝑥𝜑 → 𝜑) ∧ (∃𝑥𝜓 → 𝜓)) → ((∃𝑥𝜑 ∧ ∃𝑥𝜓) → (𝜑 ∧ 𝜓)))
5	3, 4	syl5 34	. . . . 5 ⊢ (((∃𝑥𝜑 → 𝜑) ∧ (∃𝑥𝜓 → 𝜓)) → (∃𝑥(𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜓)))
6		anim12 807	. . . . . 6 ⊢ (((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓)) → ((𝜑 ∧ 𝜓) → (∀𝑥𝜑 ∧ ∀𝑥𝜓)))
7		id 22	. . . . . . 7 ⊢ ((𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜓))
8	7	alanimi 1810	. . . . . 6 ⊢ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) → ∀𝑥(𝜑 ∧ 𝜓))
9	6, 8	syl6 35	. . . . 5 ⊢ (((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓)) → ((𝜑 ∧ 𝜓) → ∀𝑥(𝜑 ∧ 𝜓)))
10	5, 9	anim12i 611	. . . 4 ⊢ ((((∃𝑥𝜑 → 𝜑) ∧ (∃𝑥𝜓 → 𝜓)) ∧ ((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓))) → ((∃𝑥(𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜓)) ∧ ((𝜑 ∧ 𝜓) → ∀𝑥(𝜑 ∧ 𝜓))))
11	10	an4s 658	. . 3 ⊢ ((((∃𝑥𝜑 → 𝜑) ∧ (𝜑 → ∀𝑥𝜑)) ∧ ((∃𝑥𝜓 → 𝜓) ∧ (𝜓 → ∀𝑥𝜓))) → ((∃𝑥(𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜓)) ∧ ((𝜑 ∧ 𝜓) → ∀𝑥(𝜑 ∧ 𝜓))))
12	1, 2, 11	syl2anb 596	. 2 ⊢ ((Ⅎ'𝑥𝜑 ∧ Ⅎ'𝑥𝜓) → ((∃𝑥(𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜓)) ∧ ((𝜑 ∧ 𝜓) → ∀𝑥(𝜑 ∧ 𝜓))))
13		df-bj-nnf 36234	. 2 ⊢ (Ⅎ'𝑥(𝜑 ∧ 𝜓) ↔ ((∃𝑥(𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜓)) ∧ ((𝜑 ∧ 𝜓) → ∀𝑥(𝜑 ∧ 𝜓))))
14	12, 13	sylibr 233	1 ⊢ ((Ⅎ'𝑥𝜑 ∧ Ⅎ'𝑥𝜓) → Ⅎ'𝑥(𝜑 ∧ 𝜓))

Syntax hints:   → wi 4   ∧ wa 394  ∀wal 1531  ∃wex 1773  Ⅎ'wnnf 36233

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 395  df-ex 1774  df-bj-nnf 36234

This theorem is referenced by:  bj-nnfbit  36262