Metamath Proof Explorer


Theorem bj-nnfan

Description: Nonfreeness in both conjuncts implies nonfreeness in the conjunction. (Contributed by BJ, 19-Nov-2023) In classical logic, there is a proof using the definition of conjunction 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-nnfan ( ( Ⅎ' 𝑥 𝜑 ∧ Ⅎ' 𝑥 𝜓 ) → Ⅎ' 𝑥 ( 𝜑𝜓 ) )

Proof

Step Hyp Ref Expression
1 df-bj-nnf ( Ⅎ' 𝑥 𝜑 ↔ ( ( ∃ 𝑥 𝜑𝜑 ) ∧ ( 𝜑 → ∀ 𝑥 𝜑 ) ) )
2 df-bj-nnf ( Ⅎ' 𝑥 𝜓 ↔ ( ( ∃ 𝑥 𝜓𝜓 ) ∧ ( 𝜓 → ∀ 𝑥 𝜓 ) ) )
3 19.40 ( ∃ 𝑥 ( 𝜑𝜓 ) → ( ∃ 𝑥 𝜑 ∧ ∃ 𝑥 𝜓 ) )
4 anim12 ( ( ( ∃ 𝑥 𝜑𝜑 ) ∧ ( ∃ 𝑥 𝜓𝜓 ) ) → ( ( ∃ 𝑥 𝜑 ∧ ∃ 𝑥 𝜓 ) → ( 𝜑𝜓 ) ) )
5 3 4 syl5 ( ( ( ∃ 𝑥 𝜑𝜑 ) ∧ ( ∃ 𝑥 𝜓𝜓 ) ) → ( ∃ 𝑥 ( 𝜑𝜓 ) → ( 𝜑𝜓 ) ) )
6 anim12 ( ( ( 𝜑 → ∀ 𝑥 𝜑 ) ∧ ( 𝜓 → ∀ 𝑥 𝜓 ) ) → ( ( 𝜑𝜓 ) → ( ∀ 𝑥 𝜑 ∧ ∀ 𝑥 𝜓 ) ) )
7 id ( ( 𝜑𝜓 ) → ( 𝜑𝜓 ) )
8 7 alanimi ( ( ∀ 𝑥 𝜑 ∧ ∀ 𝑥 𝜓 ) → ∀ 𝑥 ( 𝜑𝜓 ) )
9 6 8 syl6 ( ( ( 𝜑 → ∀ 𝑥 𝜑 ) ∧ ( 𝜓 → ∀ 𝑥 𝜓 ) ) → ( ( 𝜑𝜓 ) → ∀ 𝑥 ( 𝜑𝜓 ) ) )
10 5 9 anim12i ( ( ( ( ∃ 𝑥 𝜑𝜑 ) ∧ ( ∃ 𝑥 𝜓𝜓 ) ) ∧ ( ( 𝜑 → ∀ 𝑥 𝜑 ) ∧ ( 𝜓 → ∀ 𝑥 𝜓 ) ) ) → ( ( ∃ 𝑥 ( 𝜑𝜓 ) → ( 𝜑𝜓 ) ) ∧ ( ( 𝜑𝜓 ) → ∀ 𝑥 ( 𝜑𝜓 ) ) ) )
11 10 an4s ( ( ( ( ∃ 𝑥 𝜑𝜑 ) ∧ ( 𝜑 → ∀ 𝑥 𝜑 ) ) ∧ ( ( ∃ 𝑥 𝜓𝜓 ) ∧ ( 𝜓 → ∀ 𝑥 𝜓 ) ) ) → ( ( ∃ 𝑥 ( 𝜑𝜓 ) → ( 𝜑𝜓 ) ) ∧ ( ( 𝜑𝜓 ) → ∀ 𝑥 ( 𝜑𝜓 ) ) ) )
12 1 2 11 syl2anb ( ( Ⅎ' 𝑥 𝜑 ∧ Ⅎ' 𝑥 𝜓 ) → ( ( ∃ 𝑥 ( 𝜑𝜓 ) → ( 𝜑𝜓 ) ) ∧ ( ( 𝜑𝜓 ) → ∀ 𝑥 ( 𝜑𝜓 ) ) ) )
13 df-bj-nnf ( Ⅎ' 𝑥 ( 𝜑𝜓 ) ↔ ( ( ∃ 𝑥 ( 𝜑𝜓 ) → ( 𝜑𝜓 ) ) ∧ ( ( 𝜑𝜓 ) → ∀ 𝑥 ( 𝜑𝜓 ) ) ) )
14 12 13 sylibr ( ( Ⅎ' 𝑥 𝜑 ∧ Ⅎ' 𝑥 𝜓 ) → Ⅎ' 𝑥 ( 𝜑𝜓 ) )