Theorem prsss 31866

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 ∧ 𝐴 ⊆ 𝐵) → ( ≤ ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))

Proof of Theorem prsss

Step	Hyp	Ref	Expression
1		ordtNEW.l	. . . . 5 ⊢ ≤ = ((le‘𝐾) ∩ (𝐵 × 𝐵))
2	1	ineq1i 4142	. . . 4 ⊢ ( ≤ ∩ (𝐴 × 𝐴)) = (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴))
3		inass 4153	. . . 4 ⊢ (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
4	2, 3	eqtri 2766	. . 3 ⊢ ( ≤ ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
5		xpss12 5604	. . . . . 6 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐴 ⊆ 𝐵) → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
6	5	anidms 567	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
7		sseqin2 4149	. . . . 5 ⊢ ((𝐴 × 𝐴) ⊆ (𝐵 × 𝐵) ↔ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
8	6, 7	sylib 217	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
9	8	ineq2d 4146	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴))) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
10	4, 9	eqtrid 2790	. 2 ⊢ (𝐴 ⊆ 𝐵 → ( ≤ ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
11	10	adantl 482	1 ⊢ ((𝐾 ∈ Proset ∧ 𝐴 ⊆ 𝐵) → ( ≤ ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))

Syntax hints:   → wi 4   ∧ wa 396   = wceq 1539   ∈ wcel 2106   ∩ cin 3886   ⊆ wss 3887   × cxp 5587  ‘cfv 6433  Basecbs 16912  lecple 16969   Proset cproset 18011

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-rab 3073  df-v 3434  df-in 3894  df-ss 3904  df-opab 5137  df-xp 5595

This theorem is referenced by:  prsssdm  31867  ordtrestNEW  31871  ordtrest2NEW  31873