Metamath Proof Explorer


Theorem resspsrbas

Description: A restricted power series algebra has the same base set. (Contributed by Mario Carneiro, 3-Jul-2015)

Ref Expression
Hypotheses resspsr.s ⊢ S = I mPwSer R
resspsr.h ⊢ H = R ↾ 𝑠 T
resspsr.u ⊢ U = I mPwSer H
resspsr.b ⊢ B = Base U
resspsr.p ⊢ P = S ↾ 𝑠 B
resspsr.2 ⊢ φ → T ∈ SubRing ⁡ R
Assertion resspsrbas ⊢ φ → B = Base P

Proof

Step Hyp Ref Expression
1 resspsr.s ⊢ S = I mPwSer R
2 resspsr.h ⊢ H = R ↾ 𝑠 T
3 resspsr.u ⊢ U = I mPwSer H
4 resspsr.b ⊢ B = Base U
5 resspsr.p ⊢ P = S ↾ 𝑠 B
6 resspsr.2 ⊢ φ → T ∈ SubRing ⁡ R
7 fvex ⊢ Base R ∈ V
8 2 subrgbas ⊢ T ∈ SubRing ⁡ R → T = Base H
9 6 8 syl ⊢ φ → T = Base H
10 eqid ⊢ Base R = Base R
11 10 subrgss ⊢ T ∈ SubRing ⁡ R → T ⊆ Base R
12 6 11 syl ⊢ φ → T ⊆ Base R
13 9 12 eqsstrrd ⊢ φ → Base H ⊆ Base R
14 13 adantr ⊢ φ ∧ I ∈ V → Base H ⊆ Base R
15 mapss ⊢ Base R ∈ V ∧ Base H ⊆ Base R → Base H f ∈ ℕ 0 I | f -1 ℕ ∈ Fin ⊆ Base R f ∈ ℕ 0 I | f -1 ℕ ∈ Fin
16 7 14 15 sylancr ⊢ φ ∧ I ∈ V → Base H f ∈ ℕ 0 I | f -1 ℕ ∈ Fin ⊆ Base R f ∈ ℕ 0 I | f -1 ℕ ∈ Fin
17 eqid ⊢ Base H = Base H
18 eqid ⊢ f ∈ ℕ 0 I | f -1 ℕ ∈ Fin = f ∈ ℕ 0 I | f -1 ℕ ∈ Fin
19 simpr ⊢ φ ∧ I ∈ V → I ∈ V
20 3 17 18 4 19 psrbas ⊢ φ ∧ I ∈ V → B = Base H f ∈ ℕ 0 I | f -1 ℕ ∈ Fin
21 eqid ⊢ Base S = Base S
22 1 10 18 21 19 psrbas ⊢ φ ∧ I ∈ V → Base S = Base R f ∈ ℕ 0 I | f -1 ℕ ∈ Fin
23 16 20 22 3sstr4d ⊢ φ ∧ I ∈ V → B ⊆ Base S
24 reldmpsr ⊢ Rel ⁡ dom ⁡ mPwSer
25 24 ovprc1 ⊢ ¬ I ∈ V → I mPwSer H = ∅
26 3 25 eqtrid ⊢ ¬ I ∈ V → U = ∅
27 26 adantl ⊢ φ ∧ ¬ I ∈ V → U = ∅
28 27 fveq2d ⊢ φ ∧ ¬ I ∈ V → Base U = Base ∅
29 base0 ⊢ ∅ = Base ∅
30 28 4 29 3eqtr4g ⊢ φ ∧ ¬ I ∈ V → B = ∅
31 0ss ⊢ ∅ ⊆ Base S
32 30 31 eqsstrdi ⊢ φ ∧ ¬ I ∈ V → B ⊆ Base S
33 23 32 pm2.61dan ⊢ φ → B ⊆ Base S
34 5 21 ressbas2 ⊢ B ⊆ Base S → B = Base P
35 33 34 syl ⊢ φ → B = Base P