Theorem bj-nnfor 37414

Description: Nonfreeness in both disjuncts implies nonfreeness in the disjunction. (Contributed by BJ, 19-Nov-2023.) In classical logic, there is a proof using the definition of disjunction in terms of implication and negation, so using bj-nnfim 37410, bj-nnfnt 37408 and bj-nnfbi 37405, but we want a proof valid in intuitionistic logic. (Proof modification is discouraged.)

Assertion

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

Proof of Theorem bj-nnfor

Step	Hyp	Ref	Expression
1		df-bj-nnf 37385	. . 3 ⊢ (Ⅎ'𝑥𝜑 ↔ ((∃𝑥𝜑 → 𝜑) ∧ (𝜑 → ∀𝑥𝜑)))
2		df-bj-nnf 37385	. . 3 ⊢ (Ⅎ'𝑥𝜓 ↔ ((∃𝑥𝜓 → 𝜓) ∧ (𝜓 → ∀𝑥𝜓)))
3		19.43 1915	. . . . . 6 ⊢ (∃𝑥(𝜑 ∨ 𝜓) ↔ (∃𝑥𝜑 ∨ ∃𝑥𝜓))
4		pm3.48 978	. . . . . 6 ⊢ (((∃𝑥𝜑 → 𝜑) ∧ (∃𝑥𝜓 → 𝜓)) → ((∃𝑥𝜑 ∨ ∃𝑥𝜓) → (𝜑 ∨ 𝜓)))
5	3, 4	biimtrid 245	. . . . 5 ⊢ (((∃𝑥𝜑 → 𝜑) ∧ (∃𝑥𝜓 → 𝜓)) → (∃𝑥(𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜓)))
6		pm3.48 978	. . . . . 6 ⊢ (((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓)) → ((𝜑 ∨ 𝜓) → (∀𝑥𝜑 ∨ ∀𝑥𝜓)))
7		19.33 1917	. . . . . 6 ⊢ ((∀𝑥𝜑 ∨ ∀𝑥𝜓) → ∀𝑥(𝜑 ∨ 𝜓))
8	6, 7	syl6 36	. . . . 5 ⊢ (((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓)) → ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓)))
9	5, 8	anim12i 625	. . . 4 ⊢ ((((∃𝑥𝜑 → 𝜑) ∧ (∃𝑥𝜓 → 𝜓)) ∧ ((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓))) → ((∃𝑥(𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜓)) ∧ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))))
10	9	an4s 673	. . 3 ⊢ ((((∃𝑥𝜑 → 𝜑) ∧ (𝜑 → ∀𝑥𝜑)) ∧ ((∃𝑥𝜓 → 𝜓) ∧ (𝜓 → ∀𝑥𝜓))) → ((∃𝑥(𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜓)) ∧ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))))
11	1, 2, 10	syl2anb 610	. 2 ⊢ ((Ⅎ'𝑥𝜑 ∧ Ⅎ'𝑥𝜓) → ((∃𝑥(𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜓)) ∧ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))))
12		df-bj-nnf 37385	. 2 ⊢ (Ⅎ'𝑥(𝜑 ∨ 𝜓) ↔ ((∃𝑥(𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜓)) ∧ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))))
13	11, 12	sylibr 237	1 ⊢ ((Ⅎ'𝑥𝜑 ∧ Ⅎ'𝑥𝜓) → Ⅎ'𝑥(𝜑 ∨ 𝜓))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∨ wo 861  ∀wal 1568  ∃wex 1812  Ⅎ'wnnf 37384

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-bj-nnf 37385

This theorem is used by: (None)