Theorem bj-nnfor 34859

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 34855, bj-nnfnt 34849 and bj-nnfbi 34834, 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 34833	. . 3 ⊢ (Ⅎ'𝑥𝜑 ↔ ((∃𝑥𝜑 → 𝜑) ∧ (𝜑 → ∀𝑥𝜑)))
2		df-bj-nnf 34833	. . 3 ⊢ (Ⅎ'𝑥𝜓 ↔ ((∃𝑥𝜓 → 𝜓) ∧ (𝜓 → ∀𝑥𝜓)))
3		19.43 1886	. . . . . 6 ⊢ (∃𝑥(𝜑 ∨ 𝜓) ↔ (∃𝑥𝜑 ∨ ∃𝑥𝜓))
4		pm3.48 960	. . . . . 6 ⊢ (((∃𝑥𝜑 → 𝜑) ∧ (∃𝑥𝜓 → 𝜓)) → ((∃𝑥𝜑 ∨ ∃𝑥𝜓) → (𝜑 ∨ 𝜓)))
5	3, 4	syl5bi 241	. . . . 5 ⊢ (((∃𝑥𝜑 → 𝜑) ∧ (∃𝑥𝜓 → 𝜓)) → (∃𝑥(𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜓)))
6		pm3.48 960	. . . . . 6 ⊢ (((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓)) → ((𝜑 ∨ 𝜓) → (∀𝑥𝜑 ∨ ∀𝑥𝜓)))
7		19.33 1888	. . . . . 6 ⊢ ((∀𝑥𝜑 ∨ ∀𝑥𝜓) → ∀𝑥(𝜑 ∨ 𝜓))
8	6, 7	syl6 35	. . . . 5 ⊢ (((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓)) → ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓)))
9	5, 8	anim12i 612	. . . 4 ⊢ ((((∃𝑥𝜑 → 𝜑) ∧ (∃𝑥𝜓 → 𝜓)) ∧ ((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓))) → ((∃𝑥(𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜓)) ∧ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))))
10	9	an4s 656	. . 3 ⊢ ((((∃𝑥𝜑 → 𝜑) ∧ (𝜑 → ∀𝑥𝜑)) ∧ ((∃𝑥𝜓 → 𝜓) ∧ (𝜓 → ∀𝑥𝜓))) → ((∃𝑥(𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜓)) ∧ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))))
11	1, 2, 10	syl2anb 597	. 2 ⊢ ((Ⅎ'𝑥𝜑 ∧ Ⅎ'𝑥𝜓) → ((∃𝑥(𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜓)) ∧ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))))
12		df-bj-nnf 34833	. 2 ⊢ (Ⅎ'𝑥(𝜑 ∨ 𝜓) ↔ ((∃𝑥(𝜑 ∨ 𝜓) → (𝜑 ∨ 𝜓)) ∧ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))))
13	11, 12	sylibr 233	1 ⊢ ((Ⅎ'𝑥𝜑 ∧ Ⅎ'𝑥𝜓) → Ⅎ'𝑥(𝜑 ∨ 𝜓))

Syntax hints:   → wi 4   ∧ wa 395   ∨ wo 843  ∀wal 1537  ∃wex 1783  Ⅎ'wnnf 34832

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-ex 1784  df-bj-nnf 34833

This theorem is referenced by: (None)