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Theorem bj-nnfor 34501
 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 34497, bj-nnfnt 34491 and bj-nnfbi 34479, but we want a proof valid in intuitionistic logic. (Proof modification is discouraged.)
Assertion
Ref Expression
bj-nnfor ((Ⅎ'𝑥𝜑 ∧ Ⅎ'𝑥𝜓) → Ⅎ'𝑥(𝜑𝜓))

Proof of Theorem bj-nnfor
StepHypRef Expression
1 df-bj-nnf 34478 . . 3 (Ⅎ'𝑥𝜑 ↔ ((∃𝑥𝜑𝜑) ∧ (𝜑 → ∀𝑥𝜑)))
2 df-bj-nnf 34478 . . 3 (Ⅎ'𝑥𝜓 ↔ ((∃𝑥𝜓𝜓) ∧ (𝜓 → ∀𝑥𝜓)))
3 19.43 1883 . . . . . 6 (∃𝑥(𝜑𝜓) ↔ (∃𝑥𝜑 ∨ ∃𝑥𝜓))
4 pm3.48 961 . . . . . 6 (((∃𝑥𝜑𝜑) ∧ (∃𝑥𝜓𝜓)) → ((∃𝑥𝜑 ∨ ∃𝑥𝜓) → (𝜑𝜓)))
53, 4syl5bi 245 . . . . 5 (((∃𝑥𝜑𝜑) ∧ (∃𝑥𝜓𝜓)) → (∃𝑥(𝜑𝜓) → (𝜑𝜓)))
6 pm3.48 961 . . . . . 6 (((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓)) → ((𝜑𝜓) → (∀𝑥𝜑 ∨ ∀𝑥𝜓)))
7 19.33 1885 . . . . . 6 ((∀𝑥𝜑 ∨ ∀𝑥𝜓) → ∀𝑥(𝜑𝜓))
86, 7syl6 35 . . . . 5 (((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓)) → ((𝜑𝜓) → ∀𝑥(𝜑𝜓)))
95, 8anim12i 615 . . . 4 ((((∃𝑥𝜑𝜑) ∧ (∃𝑥𝜓𝜓)) ∧ ((𝜑 → ∀𝑥𝜑) ∧ (𝜓 → ∀𝑥𝜓))) → ((∃𝑥(𝜑𝜓) → (𝜑𝜓)) ∧ ((𝜑𝜓) → ∀𝑥(𝜑𝜓))))
109an4s 659 . . 3 ((((∃𝑥𝜑𝜑) ∧ (𝜑 → ∀𝑥𝜑)) ∧ ((∃𝑥𝜓𝜓) ∧ (𝜓 → ∀𝑥𝜓))) → ((∃𝑥(𝜑𝜓) → (𝜑𝜓)) ∧ ((𝜑𝜓) → ∀𝑥(𝜑𝜓))))
111, 2, 10syl2anb 600 . 2 ((Ⅎ'𝑥𝜑 ∧ Ⅎ'𝑥𝜓) → ((∃𝑥(𝜑𝜓) → (𝜑𝜓)) ∧ ((𝜑𝜓) → ∀𝑥(𝜑𝜓))))
12 df-bj-nnf 34478 . 2 (Ⅎ'𝑥(𝜑𝜓) ↔ ((∃𝑥(𝜑𝜓) → (𝜑𝜓)) ∧ ((𝜑𝜓) → ∀𝑥(𝜑𝜓))))
1311, 12sylibr 237 1 ((Ⅎ'𝑥𝜑 ∧ Ⅎ'𝑥𝜓) → Ⅎ'𝑥(𝜑𝜓))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ wa 399   ∨ wo 844  ∀wal 1536  ∃wex 1781  Ⅎ'wnnf 34477 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-bj-nnf 34478 This theorem is referenced by: (None)
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