Metamath Proof Explorer


Theorem psssdm

Description: Field of a subposet. (Contributed by FL, 19-Sep-2011) (Revised by Mario Carneiro, 9-Sep-2015)

Ref Expression
Hypothesis psssdm.1 𝑋 = dom 𝑅
Assertion psssdm ( ( 𝑅 ∈ PosetRel ∧ 𝐴𝑋 ) → dom ( 𝑅 ∩ ( 𝐴 × 𝐴 ) ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 psssdm.1 𝑋 = dom 𝑅
2 1 psssdm2 ( 𝑅 ∈ PosetRel → dom ( 𝑅 ∩ ( 𝐴 × 𝐴 ) ) = ( 𝑋𝐴 ) )
3 sseqin2 ( 𝐴𝑋 ↔ ( 𝑋𝐴 ) = 𝐴 )
4 3 biimpi ( 𝐴𝑋 → ( 𝑋𝐴 ) = 𝐴 )
5 2 4 sylan9eq ( ( 𝑅 ∈ PosetRel ∧ 𝐴𝑋 ) → dom ( 𝑅 ∩ ( 𝐴 × 𝐴 ) ) = 𝐴 )