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Theorem prsss 33877
Description: Relation of a subproset. (Contributed by Thierry Arnoux, 13-Sep-2018.)
Hypotheses
Ref Expression
ordtNEW.b 𝐵 = (Base‘𝐾)
ordtNEW.l = ((le‘𝐾) ∩ (𝐵 × 𝐵))
Assertion
Ref Expression
prsss ((𝐾 ∈ Proset ∧ 𝐴𝐵) → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))

Proof of Theorem prsss
StepHypRef Expression
1 ordtNEW.l . . . . 5 = ((le‘𝐾) ∩ (𝐵 × 𝐵))
21ineq1i 4224 . . . 4 ( ∩ (𝐴 × 𝐴)) = (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴))
3 inass 4236 . . . 4 (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
42, 3eqtri 2763 . . 3 ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
5 xpss12 5704 . . . . . 6 ((𝐴𝐵𝐴𝐵) → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
65anidms 566 . . . . 5 (𝐴𝐵 → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
7 sseqin2 4231 . . . . 5 ((𝐴 × 𝐴) ⊆ (𝐵 × 𝐵) ↔ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
86, 7sylib 218 . . . 4 (𝐴𝐵 → ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
98ineq2d 4228 . . 3 (𝐴𝐵 → ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴))) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
104, 9eqtrid 2787 . 2 (𝐴𝐵 → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
1110adantl 481 1 ((𝐾 ∈ Proset ∧ 𝐴𝐵) → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1537  wcel 2106  cin 3962  wss 3963   × cxp 5687  cfv 6563  Basecbs 17245  lecple 17305   Proset cproset 18350
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-in 3970  df-ss 3980  df-opab 5211  df-xp 5695
This theorem is referenced by:  prsssdm  33878  ordtrestNEW  33882  ordtrest2NEW  33884
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