Theorem elnelall

Description: A contradiction concerning membership implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018)

	Ref	Expression
Assertion	elnelall	$$⊢A\in B\to \left(A\notin B\to \phi \right)$$

Proof

Step	Hyp	Ref	Expression
1		df-nel	$$⊢A\notin B\leftrightarrow \neg A\in B$$
2		pm2.24	$$⊢A\in B\to \left(\neg A\in B\to \phi \right)$$
3	1 2	syl5bi	$$⊢A\in B\to \left(A\notin B\to \phi \right)$$