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Metamath Proof Explorer

Theorem ax9v1

Description: First of two weakened versions of ax9v , with an extra disjoint variable condition on x , z , see comments there. (Contributed by BJ, 7-Dec-2020)

		Ref	Expression
	Assertion	ax9v1	$⊢ x = y \to (z \in x \to z \in y)$

Proof

Step	Hyp	Ref	Expression
1		ax9v	$⊢ x = y \to (z \in x \to z \in y)$