Theorem prsss 34157

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 4159	. . . 4 ⊢ ( ≤ ∩ (𝐴 × 𝐴)) = (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴))
3		inass 4170	. . . 4 ⊢ (((le‘𝐾) ∩ (𝐵 × 𝐵)) ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
4	2, 3	eqtri 2775	. . 3 ⊢ ( ≤ ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)))
5		xpss12 5651	. . . . . 6 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐴 ⊆ 𝐵) → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
6	5	anidms 573	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 × 𝐴) ⊆ (𝐵 × 𝐵))
7		sseqin2 4166	. . . . 5 ⊢ ((𝐴 × 𝐴) ⊆ (𝐵 × 𝐵) ↔ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
8	6, 7	sylib 220	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
9	8	ineq2d 4163	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((le‘𝐾) ∩ ((𝐵 × 𝐵) ∩ (𝐴 × 𝐴))) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
10	4, 9	eqtrid 2799	. 2 ⊢ (𝐴 ⊆ 𝐵 → ( ≤ ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))
11	10	adantl 484	1 ⊢ ((𝐾 ∈ Proset ∧ 𝐴 ⊆ 𝐵) → ( ≤ ∩ (𝐴 × 𝐴)) = ((le‘𝐾) ∩ (𝐴 × 𝐴)))

Syntax hints:   → wi 4   ∧ wa 398   = wceq 1550   ∈ wcel 2132   ∩ cin 3894   ⊆ wss 3895   × cxp 5634  ‘cfv 6506  Basecbs 17217  lecple 17265   Proset cproset 18296

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1805  ax-4 1819  ax-5 1920  ax-6 1977  ax-7 2018  ax-8 2134  ax-9 2142  ax-ext 2724

This theorem depends on definitions:  df-bi 209  df-an 399  df-3an 1097  df-tru 1553  df-ex 1790  df-sb 2081  df-clab 2731  df-cleq 2744  df-clel 2827  df-rab 3405  df-v 3446  df-in 3902  df-ss 3912  df-opab 5153  df-xp 5642

This theorem is referenced by:  prsssdm  34158  ordtrestNEW  34162  ordtrest2NEW  34164