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Theorem prsss 30296
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 3959 . . . 4 ( ∩ (𝐴 × 𝐴)) = (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴))
3 inass 3970 . . . 4 (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
42, 3eqtri 2792 . . 3 ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
5 xpss12 5264 . . . . . 6 ((𝐴𝐵𝐴𝐵) → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
65anidms 548 . . . . 5 (𝐴𝐵 → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
7 sseqin2 3966 . . . . 5 ((𝐴 × 𝐴) ⊆ (𝐵 × 𝐵) ↔ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
86, 7sylib 208 . . . 4 (𝐴𝐵 → ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
98ineq2d 3963 . . 3 (𝐴𝐵 → ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴))) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
104, 9syl5eq 2816 . 2 (𝐴𝐵 → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
1110adantl 467 1 ((𝐾 ∈ Preset ∧ 𝐴𝐵) → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 382   = wceq 1630  wcel 2144  cin 3720  wss 3721   × cxp 5247  cfv 6031  Basecbs 16063  lecple 16155   Preset cpreset 17133
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1869  ax-4 1884  ax-5 1990  ax-6 2056  ax-7 2092  ax-9 2153  ax-10 2173  ax-11 2189  ax-12 2202  ax-13 2407  ax-ext 2750
This theorem depends on definitions:  df-bi 197  df-an 383  df-or 827  df-tru 1633  df-ex 1852  df-nf 1857  df-sb 2049  df-clab 2757  df-cleq 2763  df-clel 2766  df-nfc 2901  df-v 3351  df-in 3728  df-ss 3735  df-opab 4845  df-xp 5255
This theorem is referenced by:  prsssdm  30297  ordtrestNEW  30301  ordtrest2NEW  30303
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