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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	\|- ( E. x E. y ph -> E. y E. x ph )

Proof

Step	Hyp	Ref	Expression
1		excom	\|- ( E. x E. y ph <-> E. y E. x ph )
2	1	biimpi	\|- ( E. x E. y ph -> E. y E. x ph )