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Metamath Proof Explorer

Theorem 19.12vvv

Description: Version of 19.12vv with a disjoint variable condition, requiring fewer axioms. See also 19.12 . (Contributed by BJ, 18-Mar-2020)

Ref Expression
Assertion 19.12vvv ( ∃ 𝑥 ∀ 𝑦 ( 𝜑 → 𝜓 ) ↔ ∀ 𝑦 ∃ 𝑥 ( 𝜑 → 𝜓 ) )

Proof

Step Hyp Ref Expression
1 19.21v ⊢ ( ∀ 𝑦 ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → ∀ 𝑦 𝜓 ) )
2 1 exbii ⊢ ( ∃ 𝑥 ∀ 𝑦 ( 𝜑 → 𝜓 ) ↔ ∃ 𝑥 ( 𝜑 → ∀ 𝑦 𝜓 ) )
3 19.36v ⊢ ( ∃ 𝑥 ( 𝜑 → ∀ 𝑦 𝜓 ) ↔ ( ∀ 𝑥 𝜑 → ∀ 𝑦 𝜓 ) )
4 19.36v ⊢ ( ∃ 𝑥 ( 𝜑 → 𝜓 ) ↔ ( ∀ 𝑥 𝜑 → 𝜓 ) )
5 4 albii ⊢ ( ∀ 𝑦 ∃ 𝑥 ( 𝜑 → 𝜓 ) ↔ ∀ 𝑦 ( ∀ 𝑥 𝜑 → 𝜓 ) )
6 19.21v ⊢ ( ∀ 𝑦 ( ∀ 𝑥 𝜑 → 𝜓 ) ↔ ( ∀ 𝑥 𝜑 → ∀ 𝑦 𝜓 ) )
7 5 6 bitr2i ⊢ ( ( ∀ 𝑥 𝜑 → ∀ 𝑦 𝜓 ) ↔ ∀ 𝑦 ∃ 𝑥 ( 𝜑 → 𝜓 ) )
8 2 3 7 3bitri ⊢ ( ∃ 𝑥 ∀ 𝑦 ( 𝜑 → 𝜓 ) ↔ ∀ 𝑦 ∃ 𝑥 ( 𝜑 → 𝜓 ) )