Metamath Proof Explorer


Theorem sspsstri

Description: Two ways of stating trichotomy with respect to inclusion. (Contributed by NM, 12-Aug-2004)

Ref Expression
Assertion sspsstri ( ( 𝐴𝐵𝐵𝐴 ) ↔ ( 𝐴𝐵𝐴 = 𝐵𝐵𝐴 ) )

Proof

Step Hyp Ref Expression
1 or32 ( ( ( 𝐴𝐵𝐵𝐴 ) ∨ 𝐴 = 𝐵 ) ↔ ( ( 𝐴𝐵𝐴 = 𝐵 ) ∨ 𝐵𝐴 ) )
2 sspss ( 𝐴𝐵 ↔ ( 𝐴𝐵𝐴 = 𝐵 ) )
3 sspss ( 𝐵𝐴 ↔ ( 𝐵𝐴𝐵 = 𝐴 ) )
4 eqcom ( 𝐵 = 𝐴𝐴 = 𝐵 )
5 4 orbi2i ( ( 𝐵𝐴𝐵 = 𝐴 ) ↔ ( 𝐵𝐴𝐴 = 𝐵 ) )
6 3 5 bitri ( 𝐵𝐴 ↔ ( 𝐵𝐴𝐴 = 𝐵 ) )
7 2 6 orbi12i ( ( 𝐴𝐵𝐵𝐴 ) ↔ ( ( 𝐴𝐵𝐴 = 𝐵 ) ∨ ( 𝐵𝐴𝐴 = 𝐵 ) ) )
8 orordir ( ( ( 𝐴𝐵𝐵𝐴 ) ∨ 𝐴 = 𝐵 ) ↔ ( ( 𝐴𝐵𝐴 = 𝐵 ) ∨ ( 𝐵𝐴𝐴 = 𝐵 ) ) )
9 7 8 bitr4i ( ( 𝐴𝐵𝐵𝐴 ) ↔ ( ( 𝐴𝐵𝐵𝐴 ) ∨ 𝐴 = 𝐵 ) )
10 df-3or ( ( 𝐴𝐵𝐴 = 𝐵𝐵𝐴 ) ↔ ( ( 𝐴𝐵𝐴 = 𝐵 ) ∨ 𝐵𝐴 ) )
11 1 9 10 3bitr4i ( ( 𝐴𝐵𝐵𝐴 ) ↔ ( 𝐴𝐵𝐴 = 𝐵𝐵𝐴 ) )