Theorem bj-nnfor

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 , bj-nnfnt and bj-nnfbi , but we want a proof valid in intuitionistic logic. (Proof modification is discouraged.)

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

Proof

Step	Hyp	Ref	Expression
1		df-bj-nnf	⊢ ( Ⅎ' 𝑥 𝜑 ↔ ( ( ∃ 𝑥 𝜑 → 𝜑 ) ∧ ( 𝜑 → ∀ 𝑥 𝜑 ) ) )
2		df-bj-nnf	⊢ ( Ⅎ' 𝑥 𝜓 ↔ ( ( ∃ 𝑥 𝜓 → 𝜓 ) ∧ ( 𝜓 → ∀ 𝑥 𝜓 ) ) )
3		19.43	⊢ ( ∃ 𝑥 ( 𝜑 ∨ 𝜓 ) ↔ ( ∃ 𝑥 𝜑 ∨ ∃ 𝑥 𝜓 ) )
4		pm3.48	⊢ ( ( ( ∃ 𝑥 𝜑 → 𝜑 ) ∧ ( ∃ 𝑥 𝜓 → 𝜓 ) ) → ( ( ∃ 𝑥 𝜑 ∨ ∃ 𝑥 𝜓 ) → ( 𝜑 ∨ 𝜓 ) ) )
5	3 4	syl5bi	⊢ ( ( ( ∃ 𝑥 𝜑 → 𝜑 ) ∧ ( ∃ 𝑥 𝜓 → 𝜓 ) ) → ( ∃ 𝑥 ( 𝜑 ∨ 𝜓 ) → ( 𝜑 ∨ 𝜓 ) ) )
6		pm3.48	⊢ ( ( ( 𝜑 → ∀ 𝑥 𝜑 ) ∧ ( 𝜓 → ∀ 𝑥 𝜓 ) ) → ( ( 𝜑 ∨ 𝜓 ) → ( ∀ 𝑥 𝜑 ∨ ∀ 𝑥 𝜓 ) ) )
7		19.33	⊢ ( ( ∀ 𝑥 𝜑 ∨ ∀ 𝑥 𝜓 ) → ∀ 𝑥 ( 𝜑 ∨ 𝜓 ) )
8	6 7	syl6	⊢ ( ( ( 𝜑 → ∀ 𝑥 𝜑 ) ∧ ( 𝜓 → ∀ 𝑥 𝜓 ) ) → ( ( 𝜑 ∨ 𝜓 ) → ∀ 𝑥 ( 𝜑 ∨ 𝜓 ) ) )
9	5 8	anim12i	⊢ ( ( ( ( ∃ 𝑥 𝜑 → 𝜑 ) ∧ ( ∃ 𝑥 𝜓 → 𝜓 ) ) ∧ ( ( 𝜑 → ∀ 𝑥 𝜑 ) ∧ ( 𝜓 → ∀ 𝑥 𝜓 ) ) ) → ( ( ∃ 𝑥 ( 𝜑 ∨ 𝜓 ) → ( 𝜑 ∨ 𝜓 ) ) ∧ ( ( 𝜑 ∨ 𝜓 ) → ∀ 𝑥 ( 𝜑 ∨ 𝜓 ) ) ) )
10	9	an4s	⊢ ( ( ( ( ∃ 𝑥 𝜑 → 𝜑 ) ∧ ( 𝜑 → ∀ 𝑥 𝜑 ) ) ∧ ( ( ∃ 𝑥 𝜓 → 𝜓 ) ∧ ( 𝜓 → ∀ 𝑥 𝜓 ) ) ) → ( ( ∃ 𝑥 ( 𝜑 ∨ 𝜓 ) → ( 𝜑 ∨ 𝜓 ) ) ∧ ( ( 𝜑 ∨ 𝜓 ) → ∀ 𝑥 ( 𝜑 ∨ 𝜓 ) ) ) )
11	1 2 10	syl2anb	⊢ ( ( Ⅎ' 𝑥 𝜑 ∧ Ⅎ' 𝑥 𝜓 ) → ( ( ∃ 𝑥 ( 𝜑 ∨ 𝜓 ) → ( 𝜑 ∨ 𝜓 ) ) ∧ ( ( 𝜑 ∨ 𝜓 ) → ∀ 𝑥 ( 𝜑 ∨ 𝜓 ) ) ) )
12		df-bj-nnf	⊢ ( Ⅎ' 𝑥 ( 𝜑 ∨ 𝜓 ) ↔ ( ( ∃ 𝑥 ( 𝜑 ∨ 𝜓 ) → ( 𝜑 ∨ 𝜓 ) ) ∧ ( ( 𝜑 ∨ 𝜓 ) → ∀ 𝑥 ( 𝜑 ∨ 𝜓 ) ) ) )
13	11 12	sylibr	⊢ ( ( Ⅎ' 𝑥 𝜑 ∧ Ⅎ' 𝑥 𝜓 ) → Ⅎ' 𝑥 ( 𝜑 ∨ 𝜓 ) )