Metamath Proof Explorer


Theorem excomim

Description: One direction of Theorem 19.11 of Margaris p. 89. (Contributed by NM, 5-Aug-1993) (Revised by Mario Carneiro, 24-Sep-2016) Remove dependencies on ax-5 , ax-6 , ax-7 , ax-10 , ax-12 . (Revised by Wolf Lammen, 8-Jan-2018)

Ref Expression
Assertion excomim ( ∃ 𝑥𝑦 𝜑 → ∃ 𝑦𝑥 𝜑 )

Proof

Step Hyp Ref Expression
1 excom ( ∃ 𝑥𝑦 𝜑 ↔ ∃ 𝑦𝑥 𝜑 )
2 1 biimpi ( ∃ 𝑥𝑦 𝜑 → ∃ 𝑦𝑥 𝜑 )