Metamath Proof Explorer


Theorem ssdif0

Description: Subclass expressed in terms of difference. Exercise 7 of TakeutiZaring p. 22. (Contributed by NM, 29-Apr-1994)

Ref Expression
Assertion ssdif0 ⊢ A ⊆ B ↔ A ∖ B = ∅

Proof

Step Hyp Ref Expression
1 iman ⊢ x ∈ A → x ∈ B ↔ ¬ x ∈ A ∧ ¬ x ∈ B
2 eldif ⊢ x ∈ A ∖ B ↔ x ∈ A ∧ ¬ x ∈ B
3 1 2 xchbinxr ⊢ x ∈ A → x ∈ B ↔ ¬ x ∈ A ∖ B
4 3 albii ⊢ ∀ x x ∈ A → x ∈ B ↔ ∀ x ¬ x ∈ A ∖ B
5 df-ss ⊢ A ⊆ B ↔ ∀ x x ∈ A → x ∈ B
6 eq0 ⊢ A ∖ B = ∅ ↔ ∀ x ¬ x ∈ A ∖ B
7 4 5 6 3bitr4i ⊢ A ⊆ B ↔ A ∖ B = ∅