Metamath Proof Explorer


Theorem pssdifcom1

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

Ref Expression
Assertion pssdifcom1 ⊢ A ⊆ C ∧ B ⊆ C → C ∖ A ⊂ B ↔ C ∖ B ⊂ A

Proof

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