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 ( ( 𝐴𝐶𝐵𝐶 ) → ( 𝐵 ⊊ ( 𝐶𝐴 ) ↔ 𝐴 ⊊ ( 𝐶𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 ssconb ( ( 𝐵𝐶𝐴𝐶 ) → ( 𝐵 ⊆ ( 𝐶𝐴 ) ↔ 𝐴 ⊆ ( 𝐶𝐵 ) ) )
2 1 ancoms ( ( 𝐴𝐶𝐵𝐶 ) → ( 𝐵 ⊆ ( 𝐶𝐴 ) ↔ 𝐴 ⊆ ( 𝐶𝐵 ) ) )
3 difcom ( ( 𝐶𝐴 ) ⊆ 𝐵 ↔ ( 𝐶𝐵 ) ⊆ 𝐴 )
4 3 notbii ( ¬ ( 𝐶𝐴 ) ⊆ 𝐵 ↔ ¬ ( 𝐶𝐵 ) ⊆ 𝐴 )
5 4 a1i ( ( 𝐴𝐶𝐵𝐶 ) → ( ¬ ( 𝐶𝐴 ) ⊆ 𝐵 ↔ ¬ ( 𝐶𝐵 ) ⊆ 𝐴 ) )
6 2 5 anbi12d ( ( 𝐴𝐶𝐵𝐶 ) → ( ( 𝐵 ⊆ ( 𝐶𝐴 ) ∧ ¬ ( 𝐶𝐴 ) ⊆ 𝐵 ) ↔ ( 𝐴 ⊆ ( 𝐶𝐵 ) ∧ ¬ ( 𝐶𝐵 ) ⊆ 𝐴 ) ) )
7 dfpss3 ( 𝐵 ⊊ ( 𝐶𝐴 ) ↔ ( 𝐵 ⊆ ( 𝐶𝐴 ) ∧ ¬ ( 𝐶𝐴 ) ⊆ 𝐵 ) )
8 dfpss3 ( 𝐴 ⊊ ( 𝐶𝐵 ) ↔ ( 𝐴 ⊆ ( 𝐶𝐵 ) ∧ ¬ ( 𝐶𝐵 ) ⊆ 𝐴 ) )
9 6 7 8 3bitr4g ( ( 𝐴𝐶𝐵𝐶 ) → ( 𝐵 ⊊ ( 𝐶𝐴 ) ↔ 𝐴 ⊊ ( 𝐶𝐵 ) ) )