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Metamath Proof Explorer

Theorem 19.12vv

Description: Special case of 19.12 where its converse holds. See 19.12vvv for a version with a disjoint variable condition requiring fewer axioms. (Contributed by NM, 18-Jul-2001) (Revised by Andrew Salmon, 11-Jul-2011)

Ref Expression
Assertion 19.12vv ( ∃ 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 nfv ⊢ Ⅎ 𝑥 𝜓
4 3 nfal ⊢ Ⅎ 𝑥 ∀ 𝑦 𝜓
5 4 19.36 ⊢ ( ∃ 𝑥 ( 𝜑 → ∀ 𝑦 𝜓 ) ↔ ( ∀ 𝑥 𝜑 → ∀ 𝑦 𝜓 ) )
6 19.36v ⊢ ( ∃ 𝑥 ( 𝜑 → 𝜓 ) ↔ ( ∀ 𝑥 𝜑 → 𝜓 ) )
7 6 albii ⊢ ( ∀ 𝑦 ∃ 𝑥 ( 𝜑 → 𝜓 ) ↔ ∀ 𝑦 ( ∀ 𝑥 𝜑 → 𝜓 ) )
8 nfv ⊢ Ⅎ 𝑦 𝜑
9 8 nfal ⊢ Ⅎ 𝑦 ∀ 𝑥 𝜑
10 9 19.21 ⊢ ( ∀ 𝑦 ( ∀ 𝑥 𝜑 → 𝜓 ) ↔ ( ∀ 𝑥 𝜑 → ∀ 𝑦 𝜓 ) )
11 7 10 bitr2i ⊢ ( ( ∀ 𝑥 𝜑 → ∀ 𝑦 𝜓 ) ↔ ∀ 𝑦 ∃ 𝑥 ( 𝜑 → 𝜓 ) )
12 2 5 11 3bitri ⊢ ( ∃ 𝑥 ∀ 𝑦 ( 𝜑 → 𝜓 ) ↔ ∀ 𝑦 ∃ 𝑥 ( 𝜑 → 𝜓 ) )