Theorem dfss6

Description: Alternate definition of subclass relationship. (Contributed by RP, 16-Apr-2020)

	Ref	Expression
Assertion	dfss6	⊢ ( 𝐴 ⊆ 𝐵 ↔ ¬ ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		dfss2	⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
2		notnotb	⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ↔ ¬ ¬ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
3	1 2	bitri	⊢ ( 𝐴 ⊆ 𝐵 ↔ ¬ ¬ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
4		exanali	⊢ ( ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵 ) ↔ ¬ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
5	3 4	xchbinxr	⊢ ( 𝐴 ⊆ 𝐵 ↔ ¬ ∃ 𝑥 ( 𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵 ) )