Theorem nssrex

Description: Negation of subclass relationship. (Contributed by Glauco Siliprandi, 3-Mar-2021)

	Ref	Expression
Assertion	nssrex	⊢ ( ¬ 𝐴 ⊆ 𝐵 ↔ ∃ 𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		nss	⊢ ( ¬ 𝐴 ⊆ 𝐵 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵 ) )
2		df-rex	⊢ ( ∃ 𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 ↔ ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵 ) )
3	1 2	bitr4i	⊢ ( ¬ 𝐴 ⊆ 𝐵 ↔ ∃ 𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 )