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Theorem prsss 29947
Description: Relation of a subpreset. (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 ∈ Preset ∧ 𝐴𝐵) → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))

Proof of Theorem prsss
StepHypRef Expression
1 ordtNEW.l . . . . 5 = ((le‘𝐾) ∩ (𝐵 × 𝐵))
21ineq1i 3808 . . . 4 ( ∩ (𝐴 × 𝐴)) = (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴))
3 inass 3821 . . . 4 (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
42, 3eqtri 2643 . . 3 ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
5 xpss12 5223 . . . . . 6 ((𝐴𝐵𝐴𝐵) → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
65anidms 677 . . . . 5 (𝐴𝐵 → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
7 sseqin2 3815 . . . . 5 ((𝐴 × 𝐴) ⊆ (𝐵 × 𝐵) ↔ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
86, 7sylib 208 . . . 4 (𝐴𝐵 → ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
98ineq2d 3812 . . 3 (𝐴𝐵 → ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴))) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
104, 9syl5eq 2667 . 2 (𝐴𝐵 → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
1110adantl 482 1 ((𝐾 ∈ Preset ∧ 𝐴𝐵) → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 384   = wceq 1482  wcel 1989  cin 3571  wss 3572   × cxp 5110  cfv 5886  Basecbs 15851  lecple 15942   Preset cpreset 16920
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1721  ax-4 1736  ax-5 1838  ax-6 1887  ax-7 1934  ax-9 1998  ax-10 2018  ax-11 2033  ax-12 2046  ax-13 2245  ax-ext 2601
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1485  df-ex 1704  df-nf 1709  df-sb 1880  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2752  df-v 3200  df-in 3579  df-ss 3586  df-opab 4711  df-xp 5118
This theorem is referenced by:  prsssdm  29948  ordtrestNEW  29952  ordtrest2NEW  29954
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