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		df-ss	⊢ ( 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 ) )