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Theorem prsss 31451
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 4109 . . . 4 ( ∩ (𝐴 × 𝐴)) = (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴))
3 inass 4120 . . . 4 (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
42, 3eqtri 2762 . . 3 ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
5 xpss12 5550 . . . . . 6 ((𝐴𝐵𝐴𝐵) → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
65anidms 570 . . . . 5 (𝐴𝐵 → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
7 sseqin2 4116 . . . . 5 ((𝐴 × 𝐴) ⊆ (𝐵 × 𝐵) ↔ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
86, 7sylib 221 . . . 4 (𝐴𝐵 → ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
98ineq2d 4113 . . 3 (𝐴𝐵 → ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴))) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
104, 9syl5eq 2786 . 2 (𝐴𝐵 → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
1110adantl 485 1 ((𝐾 ∈ Proset ∧ 𝐴𝐵) → ( ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 399   = wceq 1542  wcel 2114  cin 3852  wss 3853   × cxp 5533  cfv 6350  Basecbs 16599  lecple 16688   Proset cproset 17665
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-ext 2711
This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1545  df-ex 1787  df-sb 2075  df-clab 2718  df-cleq 2731  df-clel 2812  df-rab 3063  df-v 3402  df-in 3860  df-ss 3870  df-opab 5103  df-xp 5541
This theorem is referenced by:  prsssdm  31452  ordtrestNEW  31456  ordtrest2NEW  31458
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