Metamath Proof Explorer


Theorem sscon

Description: Contraposition law for subsets. Exercise 15 of TakeutiZaring p. 22. (Contributed by NM, 22-Mar-1998)

Ref Expression
Assertion sscon ⊢ A ⊆ B → C ∖ B ⊆ C ∖ A

Proof

Step Hyp Ref Expression
1 ssel ⊢ A ⊆ B → x ∈ A → x ∈ B
2 1 con3d ⊢ A ⊆ B → ¬ x ∈ B → ¬ x ∈ A
3 2 anim2d ⊢ A ⊆ B → x ∈ C ∧ ¬ x ∈ B → x ∈ C ∧ ¬ x ∈ A
4 eldif ⊢ x ∈ C ∖ B ↔ x ∈ C ∧ ¬ x ∈ B
5 eldif ⊢ x ∈ C ∖ A ↔ x ∈ C ∧ ¬ x ∈ A
6 3 4 5 3imtr4g ⊢ A ⊆ B → x ∈ C ∖ B → x ∈ C ∖ A
7 6 ssrdv ⊢ A ⊆ B → C ∖ B ⊆ C ∖ A