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Metamath Proof Explorer

Theorem 3eltr4g

Description: Substitution of equal classes into membership relation. (Contributed by Mario Carneiro, 6-Jan-2017) (Proof shortened by Wolf Lammen, 23-Nov-2019)

	Ref	Expression
Hypotheses	3eltr4g.1	\|- ( ph -> A e. B )
	3eltr4g.2	\|- C = A
	3eltr4g.3	\|- D = B
Assertion	3eltr4g	\|- ( ph -> C e. D )

Proof

Step	Hyp	Ref	Expression
1		3eltr4g.1	\|- ( ph -> A e. B )
2		3eltr4g.2	\|- C = A
3		3eltr4g.3	\|- D = B
4	2 1	eqeltrid	\|- ( ph -> C e. B )
5	4 3	eleqtrrdi	\|- ( ph -> C e. D )