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Theorem prsss 31058
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 4182 . . . 4 ( ∩ (𝐴 × 𝐴)) = (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴))
3 inass 4193 . . . 4 (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
42, 3eqtri 2841 . . 3 ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
5 xpss12 5563 . . . . . 6 ((𝐴𝐵𝐴𝐵) → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
65anidms 567 . . . . 5 (𝐴𝐵 → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
7 sseqin2 4189 . . . . 5 ((𝐴 × 𝐴) ⊆ (𝐵 × 𝐵) ↔ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
86, 7sylib 219 . . . 4 (𝐴𝐵 → ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
98ineq2d 4186 . . 3 (𝐴𝐵 → ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴))) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
104, 9syl5eq 2865 . 2 (𝐴𝐵 → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
1110adantl 482 1 ((𝐾 ∈ Proset ∧ 𝐴𝐵) → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 396   = wceq 1528  wcel 2105  cin 3932  wss 3933   × cxp 5546  cfv 6348  Basecbs 16471  lecple 16560   Proset cproset 17524
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2151  ax-12 2167  ax-ext 2790
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-tru 1531  df-ex 1772  df-nf 1776  df-sb 2061  df-clab 2797  df-cleq 2811  df-clel 2890  df-nfc 2960  df-rab 3144  df-v 3494  df-in 3940  df-ss 3949  df-opab 5120  df-xp 5554
This theorem is referenced by:  prsssdm  31059  ordtrestNEW  31063  ordtrest2NEW  31065
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