This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

Theorem nssss

Description: Negation of subclass relationship. Compare nss . (Contributed by NM, 30-Jun-2004) (Proof shortened by Andrew Salmon, 25-Jul-2011)

		Ref	Expression
	Assertion	nssss	$⊢ \neg A \subseteq B \leftrightarrow \exists x (x \subseteq A \land \neg x \subseteq B)$

Step	Hyp	Ref	Expression
1		exanali	$⊢ \exists x (x \subseteq A \land \neg x \subseteq B) \leftrightarrow \neg \forall x (x \subseteq A \to x \subseteq B)$
2		ssextss	$⊢ A \subseteq B \leftrightarrow \forall x (x \subseteq A \to x \subseteq B)$
3	1 2	xchbinxr	$⊢ \exists x (x \subseteq A \land \neg x \subseteq B) \leftrightarrow \neg A \subseteq B$
4	3	bicomi	$⊢ \neg A \subseteq B \leftrightarrow \exists x (x \subseteq A \land \neg x \subseteq B)$