Metamath Proof Explorer


Theorem nss

Description: Negation of subclass relationship. Exercise 13 of TakeutiZaring p. 18. (Contributed by NM, 25-Feb-1996) (Proof shortened by Andrew Salmon, 21-Jun-2011)

Ref Expression
Assertion nss ¬ A B x x A ¬ x B

Proof

Step Hyp Ref Expression
1 exanali x x A ¬ x B ¬ x x A x B
2 dfss2 A B x x A x B
3 1 2 xchbinxr x x A ¬ x B ¬ A B
4 3 bicomi ¬ A B x x A ¬ x B