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Metamath Proof Explorer

Theorem minel

Description: A minimum element of a class has no elements in common with the class. (Contributed by NM, 22-Jun-1994) (Proof shortened by JJ, 14-Jul-2021)

		Ref	Expression
	Assertion	minel	$⊢ (A \in B \land C \cap B = \emptyset) \to \neg A \in C$

Proof

Step	Hyp	Ref	Expression
1		inelcm	$⊢ (A \in C \land A \in B) \to C \cap B \neq \emptyset$
2	1	expcom	$⊢ A \in B \to (A \in C \to C \cap B \neq \emptyset)$
3	2	necon2bd	$⊢ A \in B \to (C \cap B = \emptyset \to \neg A \in C)$
4	3	imp	$⊢ (A \in B \land C \cap B = \emptyset) \to \neg A \in C$