Theorem trrelsuperreldg 44814

Description: Concrete construction of a superclass of relation 𝑅 which is a transitive relation. (Contributed by RP, 25-Dec-2019.)

Hypotheses

Ref	Expression
trrelsuperreldg.r	⊢ (𝜑 → Rel 𝑅)
trrelsuperreldg.s	⊢ (𝜑 → 𝑆 = (dom 𝑅 × ran 𝑅))

Assertion

Ref	Expression
trrelsuperreldg	⊢ (𝜑 → (𝑅 ⊆ 𝑆 ∧ (𝑆 ∘ 𝑆) ⊆ 𝑆))

Proof of Theorem trrelsuperreldg

Step	Hyp	Ref	Expression
1		trrelsuperreldg.r	. . . 4 ⊢ (𝜑 → Rel 𝑅)
2		relssdmrn 6278	. . . 4 ⊢ (Rel 𝑅 → 𝑅 ⊆ (dom 𝑅 × ran 𝑅))
3	1, 2	syl 18	. . 3 ⊢ (𝜑 → 𝑅 ⊆ (dom 𝑅 × ran 𝑅))
4		trrelsuperreldg.s	. . 3 ⊢ (𝜑 → 𝑆 = (dom 𝑅 × ran 𝑅))
5	3, 4	sseqtrrd 3950	. 2 ⊢ (𝜑 → 𝑅 ⊆ 𝑆)
6		xptrrel 15148	. . . 4 ⊢ ((dom 𝑅 × ran 𝑅) ∘ (dom 𝑅 × ran 𝑅)) ⊆ (dom 𝑅 × ran 𝑅)
7	6	a1i 11	. . 3 ⊢ (𝜑 → ((dom 𝑅 × ran 𝑅) ∘ (dom 𝑅 × ran 𝑅)) ⊆ (dom 𝑅 × ran 𝑅))
8	4, 4	coeq12d 5847	. . 3 ⊢ (𝜑 → (𝑆 ∘ 𝑆) = ((dom 𝑅 × ran 𝑅) ∘ (dom 𝑅 × ran 𝑅)))
9	7, 8, 4	3sstr4d 3968	. 2 ⊢ (𝜑 → (𝑆 ∘ 𝑆) ⊆ 𝑆)
10	5, 9	jca 521	1 ⊢ (𝜑 → (𝑅 ⊆ 𝑆 ∧ (𝑆 ∘ 𝑆) ⊆ 𝑆))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1569   ⊆ wss 3900   × cxp 5653  dom cdm 5655  ran crn 5656   ∘ ccom 5659  Rel wrel 5660

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735  ax-sep 5252  ax-pr 5395

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5171  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667

This theorem is used by: (None)