Metamath Proof Explorer


Theorem psssdm2

Description: Field of a subposet. (Contributed by Mario Carneiro, 9-Sep-2015)

Ref Expression
Hypothesis psssdm.1 ⊢ X = dom ⁡ R
Assertion psssdm2 ⊢ R ∈ PosetRel → dom ⁡ R ∩ A × A = X ∩ A

Proof

Step Hyp Ref Expression
1 psssdm.1 ⊢ X = dom ⁡ R
2 dmin ⊢ dom ⁡ R ∩ A × A ⊆ dom ⁡ R ∩ dom ⁡ A × A
3 1 eqcomi ⊢ dom ⁡ R = X
4 dmxpid ⊢ dom ⁡ A × A = A
5 3 4 ineq12i ⊢ dom ⁡ R ∩ dom ⁡ A × A = X ∩ A
6 2 5 sseqtri ⊢ dom ⁡ R ∩ A × A ⊆ X ∩ A
7 6 a1i ⊢ R ∈ PosetRel → dom ⁡ R ∩ A × A ⊆ X ∩ A
8 simpr ⊢ R ∈ PosetRel ∧ x ∈ X ∩ A → x ∈ X ∩ A
9 8 elin2d ⊢ R ∈ PosetRel ∧ x ∈ X ∩ A → x ∈ A
10 elinel1 ⊢ x ∈ X ∩ A → x ∈ X
11 1 psref ⊢ R ∈ PosetRel ∧ x ∈ X → x R x
12 10 11 sylan2 ⊢ R ∈ PosetRel ∧ x ∈ X ∩ A → x R x
13 brinxp2 ⊢ x R ∩ A × A x ↔ x ∈ A ∧ x ∈ A ∧ x R x
14 9 9 12 13 syl21anbrc ⊢ R ∈ PosetRel ∧ x ∈ X ∩ A → x R ∩ A × A x
15 vex ⊢ x ∈ V
16 15 15 breldm ⊢ x R ∩ A × A x → x ∈ dom ⁡ R ∩ A × A
17 14 16 syl ⊢ R ∈ PosetRel ∧ x ∈ X ∩ A → x ∈ dom ⁡ R ∩ A × A
18 7 17 eqelssd ⊢ R ∈ PosetRel → dom ⁡ R ∩ A × A = X ∩ A