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	⊢ ( ¬ 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		ssextss	⊢ ( 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 ) )