Metamath Proof Explorer


Theorem acsmap2d

Description: In an algebraic closure system, if S and T have the same closure and S is independent, then there is a map f from T into the set of finite subsets of S such that S equals the union of ran f . This is proven by taking the map f from acsmapd and observing that, since S and T have the same closure, the closure of U. ran f must contain S . Since S is independent, by mrissmrcd , U. ran f must equal S . See Section II.5 in Cohn p. 81 to 82. (Contributed by David Moews, 1-May-2017)

Ref Expression
Hypotheses acsmap2d.1 ⊢ φ → A ∈ ACS ⁡ X
acsmap2d.2 ⊢ N = mrCls ⁡ A
acsmap2d.3 ⊢ I = mrInd ⁡ A
acsmap2d.4 ⊢ φ → S ∈ I
acsmap2d.5 ⊢ φ → T ⊆ X
acsmap2d.6 ⊢ φ → N ⁡ S = N ⁡ T
Assertion acsmap2d ⊢ φ → ∃ f f : T ⟶ 𝒫 S ∩ Fin ∧ S = ⋃ ran ⁡ f

Proof

Step Hyp Ref Expression
1 acsmap2d.1 ⊢ φ → A ∈ ACS ⁡ X
2 acsmap2d.2 ⊢ N = mrCls ⁡ A
3 acsmap2d.3 ⊢ I = mrInd ⁡ A
4 acsmap2d.4 ⊢ φ → S ∈ I
5 acsmap2d.5 ⊢ φ → T ⊆ X
6 acsmap2d.6 ⊢ φ → N ⁡ S = N ⁡ T
7 1 acsmred ⊢ φ → A ∈ Moore ⁡ X
8 3 7 4 mrissd ⊢ φ → S ⊆ X
9 7 2 5 mrcssidd ⊢ φ → T ⊆ N ⁡ T
10 9 6 sseqtrrd ⊢ φ → T ⊆ N ⁡ S
11 1 2 8 10 acsmapd ⊢ φ → ∃ f f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f
12 simprl ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → f : T ⟶ 𝒫 S ∩ Fin
13 7 adantr ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → A ∈ Moore ⁡ X
14 4 adantr ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → S ∈ I
15 3 13 14 mrissd ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → S ⊆ X
16 13 2 15 mrcssidd ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → S ⊆ N ⁡ S
17 6 adantr ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → N ⁡ S = N ⁡ T
18 simprr ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → T ⊆ N ⁡ ⋃ ran ⁡ f
19 13 2 mrcssvd ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → N ⁡ ⋃ ran ⁡ f ⊆ X
20 13 2 18 19 mrcssd ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → N ⁡ T ⊆ N ⁡ N ⁡ ⋃ ran ⁡ f
21 frn ⊢ f : T ⟶ 𝒫 S ∩ Fin → ran ⁡ f ⊆ 𝒫 S ∩ Fin
22 21 unissd ⊢ f : T ⟶ 𝒫 S ∩ Fin → ⋃ ran ⁡ f ⊆ ⋃ 𝒫 S ∩ Fin
23 unifpw ⊢ ⋃ 𝒫 S ∩ Fin = S
24 22 23 sseqtrdi ⊢ f : T ⟶ 𝒫 S ∩ Fin → ⋃ ran ⁡ f ⊆ S
25 24 ad2antrl ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → ⋃ ran ⁡ f ⊆ S
26 25 15 sstrd ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → ⋃ ran ⁡ f ⊆ X
27 13 2 26 mrcidmd ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → N ⁡ N ⁡ ⋃ ran ⁡ f = N ⁡ ⋃ ran ⁡ f
28 20 27 sseqtrd ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → N ⁡ T ⊆ N ⁡ ⋃ ran ⁡ f
29 17 28 eqsstrd ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → N ⁡ S ⊆ N ⁡ ⋃ ran ⁡ f
30 16 29 sstrd ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → S ⊆ N ⁡ ⋃ ran ⁡ f
31 13 2 3 30 25 14 mrissmrcd ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → S = ⋃ ran ⁡ f
32 12 31 jca ⊢ φ ∧ f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → f : T ⟶ 𝒫 S ∩ Fin ∧ S = ⋃ ran ⁡ f
33 32 ex ⊢ φ → f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → f : T ⟶ 𝒫 S ∩ Fin ∧ S = ⋃ ran ⁡ f
34 33 eximdv ⊢ φ → ∃ f f : T ⟶ 𝒫 S ∩ Fin ∧ T ⊆ N ⁡ ⋃ ran ⁡ f → ∃ f f : T ⟶ 𝒫 S ∩ Fin ∧ S = ⋃ ran ⁡ f
35 11 34 mpd ⊢ φ → ∃ f f : T ⟶ 𝒫 S ∩ Fin ∧ S = ⋃ ran ⁡ f