Metamath Proof Explorer


Theorem isps

Description: The predicate "is a poset" i.e. a transitive, reflexive, antisymmetric relation. (Contributed by NM, 11-May-2008)

Ref Expression
Assertion isps ( 𝑅𝐴 → ( 𝑅 ∈ PosetRel ↔ ( Rel 𝑅 ∧ ( 𝑅𝑅 ) ⊆ 𝑅 ∧ ( 𝑅 𝑅 ) = ( I ↾ 𝑅 ) ) ) )

Proof

Step Hyp Ref Expression
1 releq ( 𝑟 = 𝑅 → ( Rel 𝑟 ↔ Rel 𝑅 ) )
2 coeq1 ( 𝑟 = 𝑅 → ( 𝑟𝑟 ) = ( 𝑅𝑟 ) )
3 coeq2 ( 𝑟 = 𝑅 → ( 𝑅𝑟 ) = ( 𝑅𝑅 ) )
4 2 3 eqtrd ( 𝑟 = 𝑅 → ( 𝑟𝑟 ) = ( 𝑅𝑅 ) )
5 id ( 𝑟 = 𝑅𝑟 = 𝑅 )
6 4 5 sseq12d ( 𝑟 = 𝑅 → ( ( 𝑟𝑟 ) ⊆ 𝑟 ↔ ( 𝑅𝑅 ) ⊆ 𝑅 ) )
7 cnveq ( 𝑟 = 𝑅 𝑟 = 𝑅 )
8 5 7 ineq12d ( 𝑟 = 𝑅 → ( 𝑟 𝑟 ) = ( 𝑅 𝑅 ) )
9 unieq ( 𝑟 = 𝑅 𝑟 = 𝑅 )
10 9 unieqd ( 𝑟 = 𝑅 𝑟 = 𝑅 )
11 10 reseq2d ( 𝑟 = 𝑅 → ( I ↾ 𝑟 ) = ( I ↾ 𝑅 ) )
12 8 11 eqeq12d ( 𝑟 = 𝑅 → ( ( 𝑟 𝑟 ) = ( I ↾ 𝑟 ) ↔ ( 𝑅 𝑅 ) = ( I ↾ 𝑅 ) ) )
13 1 6 12 3anbi123d ( 𝑟 = 𝑅 → ( ( Rel 𝑟 ∧ ( 𝑟𝑟 ) ⊆ 𝑟 ∧ ( 𝑟 𝑟 ) = ( I ↾ 𝑟 ) ) ↔ ( Rel 𝑅 ∧ ( 𝑅𝑅 ) ⊆ 𝑅 ∧ ( 𝑅 𝑅 ) = ( I ↾ 𝑅 ) ) ) )
14 df-ps PosetRel = { 𝑟 ∣ ( Rel 𝑟 ∧ ( 𝑟𝑟 ) ⊆ 𝑟 ∧ ( 𝑟 𝑟 ) = ( I ↾ 𝑟 ) ) }
15 13 14 elab2g ( 𝑅𝐴 → ( 𝑅 ∈ PosetRel ↔ ( Rel 𝑅 ∧ ( 𝑅𝑅 ) ⊆ 𝑅 ∧ ( 𝑅 𝑅 ) = ( I ↾ 𝑅 ) ) ) )