Metamath Proof Explorer


Theorem pssdifcom2

Description: Two ways to express non-covering pairs of subsets. (Contributed by Stefan O'Rear, 31-Oct-2014)

Ref Expression
Assertion pssdifcom2 ⊢ A ⊆ C ∧ B ⊆ C → B ⊂ C ∖ A ↔ A ⊂ C ∖ B

Proof

Step Hyp Ref Expression
1 ssconb ⊢ B ⊆ C ∧ A ⊆ C → B ⊆ C ∖ A ↔ A ⊆ C ∖ B
2 1 ancoms ⊢ A ⊆ C ∧ B ⊆ C → B ⊆ C ∖ A ↔ A ⊆ C ∖ B
3 difcom ⊢ C ∖ A ⊆ B ↔ C ∖ B ⊆ A
4 3 notbii ⊢ ¬ C ∖ A ⊆ B ↔ ¬ C ∖ B ⊆ A
5 4 a1i ⊢ A ⊆ C ∧ B ⊆ C → ¬ C ∖ A ⊆ B ↔ ¬ C ∖ B ⊆ A
6 2 5 anbi12d ⊢ A ⊆ C ∧ B ⊆ C → B ⊆ C ∖ A ∧ ¬ C ∖ A ⊆ B ↔ A ⊆ C ∖ B ∧ ¬ C ∖ B ⊆ A
7 dfpss3 ⊢ B ⊂ C ∖ A ↔ B ⊆ C ∖ A ∧ ¬ C ∖ A ⊆ B
8 dfpss3 ⊢ A ⊂ C ∖ B ↔ A ⊆ C ∖ B ∧ ¬ C ∖ B ⊆ A
9 6 7 8 3bitr4g ⊢ A ⊆ C ∧ B ⊆ C → B ⊂ C ∖ A ↔ A ⊂ C ∖ B