Theorem bj-nnford 34670

Description: Nonfreeness in both disjuncts implies nonfreeness in the disjunction, deduction form. See comments for bj-nnfor 34669 and bj-nnfand 34668. (Contributed by BJ, 2-Dec-2023.) (Proof modification is discouraged.)

Hypotheses

Ref	Expression
bj-nnford.1	⊢ (𝜑 → Ⅎ'𝑥𝜓)
bj-nnford.2	⊢ (𝜑 → Ⅎ'𝑥𝜒)

Assertion

Ref	Expression
bj-nnford	⊢ (𝜑 → Ⅎ'𝑥(𝜓 ∨ 𝜒))

Proof of Theorem bj-nnford

Step	Hyp	Ref	Expression
1		19.43 1890	. . 3 ⊢ (∃𝑥(𝜓 ∨ 𝜒) ↔ (∃𝑥𝜓 ∨ ∃𝑥𝜒))
2		bj-nnford.1	. . . . 5 ⊢ (𝜑 → Ⅎ'𝑥𝜓)
3	2	bj-nnfed 34651	. . . 4 ⊢ (𝜑 → (∃𝑥𝜓 → 𝜓))
4		bj-nnford.2	. . . . 5 ⊢ (𝜑 → Ⅎ'𝑥𝜒)
5	4	bj-nnfed 34651	. . . 4 ⊢ (𝜑 → (∃𝑥𝜒 → 𝜒))
6	3, 5	orim12d 965	. . 3 ⊢ (𝜑 → ((∃𝑥𝜓 ∨ ∃𝑥𝜒) → (𝜓 ∨ 𝜒)))
7	1, 6	syl5bi 245	. 2 ⊢ (𝜑 → (∃𝑥(𝜓 ∨ 𝜒) → (𝜓 ∨ 𝜒)))
8	2	bj-nnfad 34648	. . . 4 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))
9	4	bj-nnfad 34648	. . . 4 ⊢ (𝜑 → (𝜒 → ∀𝑥𝜒))
10	8, 9	orim12d 965	. . 3 ⊢ (𝜑 → ((𝜓 ∨ 𝜒) → (∀𝑥𝜓 ∨ ∀𝑥𝜒)))
11		19.33 1892	. . 3 ⊢ ((∀𝑥𝜓 ∨ ∀𝑥𝜒) → ∀𝑥(𝜓 ∨ 𝜒))
12	10, 11	syl6 35	. 2 ⊢ (𝜑 → ((𝜓 ∨ 𝜒) → ∀𝑥(𝜓 ∨ 𝜒)))
13		df-bj-nnf 34643	. 2 ⊢ (Ⅎ'𝑥(𝜓 ∨ 𝜒) ↔ ((∃𝑥(𝜓 ∨ 𝜒) → (𝜓 ∨ 𝜒)) ∧ ((𝜓 ∨ 𝜒) → ∀𝑥(𝜓 ∨ 𝜒))))
14	7, 12, 13	sylanbrc 586	1 ⊢ (𝜑 → Ⅎ'𝑥(𝜓 ∨ 𝜒))

Syntax hints:   → wi 4   ∨ wo 847  ∀wal 1541  ∃wex 1787  Ⅎ'wnnf 34642

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-ex 1788  df-bj-nnf 34643

This theorem is referenced by: (None)