Proof of Theorem mendvscafval
Step | Hyp | Ref
| Expression |
1 | | mendvscafval.a |
. . 3
⊢ 𝐴 = (MEndo‘𝑀) |
2 | 1 | fveq2i 6759 |
. 2
⊢ (
·𝑠 ‘𝐴) = ( ·𝑠
‘(MEndo‘𝑀)) |
3 | | mendvscafval.b |
. . . . . . 7
⊢ 𝐵 = (Base‘𝐴) |
4 | 1 | mendbas 40925 |
. . . . . . 7
⊢ (𝑀 LMHom 𝑀) = (Base‘𝐴) |
5 | 3, 4 | eqtr4i 2769 |
. . . . . 6
⊢ 𝐵 = (𝑀 LMHom 𝑀) |
6 | | eqid 2738 |
. . . . . 6
⊢ (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘f
(+g‘𝑀)𝑦)) = (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘f
(+g‘𝑀)𝑦)) |
7 | | eqid 2738 |
. . . . . 6
⊢ (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘ 𝑦)) = (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘ 𝑦)) |
8 | | mendvscafval.s |
. . . . . 6
⊢ 𝑆 = (Scalar‘𝑀) |
9 | | mendvscafval.k |
. . . . . . 7
⊢ 𝐾 = (Base‘𝑆) |
10 | | eqid 2738 |
. . . . . . 7
⊢ 𝐵 = 𝐵 |
11 | | mendvscafval.e |
. . . . . . . . 9
⊢ 𝐸 = (Base‘𝑀) |
12 | 11 | xpeq1i 5606 |
. . . . . . . 8
⊢ (𝐸 × {𝑥}) = ((Base‘𝑀) × {𝑥}) |
13 | | eqid 2738 |
. . . . . . . 8
⊢ 𝑦 = 𝑦 |
14 | | mendvscafval.v |
. . . . . . . . 9
⊢ · = (
·𝑠 ‘𝑀) |
15 | | ofeq 7514 |
. . . . . . . . 9
⊢ ( · = (
·𝑠 ‘𝑀) → ∘f · =
∘f ( ·𝑠 ‘𝑀)) |
16 | 14, 15 | ax-mp 5 |
. . . . . . . 8
⊢
∘f · =
∘f ( ·𝑠 ‘𝑀) |
17 | 12, 13, 16 | oveq123i 7269 |
. . . . . . 7
⊢ ((𝐸 × {𝑥}) ∘f · 𝑦) = (((Base‘𝑀) × {𝑥}) ∘f (
·𝑠 ‘𝑀)𝑦) |
18 | 9, 10, 17 | mpoeq123i 7329 |
. . . . . 6
⊢ (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦)) = (𝑥 ∈ (Base‘𝑆), 𝑦 ∈ 𝐵 ↦ (((Base‘𝑀) × {𝑥}) ∘f (
·𝑠 ‘𝑀)𝑦)) |
19 | 5, 6, 7, 8, 18 | mendval 40924 |
. . . . 5
⊢ (𝑀 ∈ V →
(MEndo‘𝑀) =
({〈(Base‘ndx), 𝐵〉, 〈(+g‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘f
(+g‘𝑀)𝑦))〉, 〈(.r‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘ 𝑦))〉} ∪ {〈(Scalar‘ndx),
𝑆〉, 〈(
·𝑠 ‘ndx), (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦))〉})) |
20 | 19 | fveq2d 6760 |
. . . 4
⊢ (𝑀 ∈ V → (
·𝑠 ‘(MEndo‘𝑀)) = ( ·𝑠
‘({〈(Base‘ndx), 𝐵〉, 〈(+g‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘f
(+g‘𝑀)𝑦))〉, 〈(.r‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘ 𝑦))〉} ∪ {〈(Scalar‘ndx),
𝑆〉, 〈(
·𝑠 ‘ndx), (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦))〉}))) |
21 | 9 | fvexi 6770 |
. . . . . 6
⊢ 𝐾 ∈ V |
22 | 3 | fvexi 6770 |
. . . . . 6
⊢ 𝐵 ∈ V |
23 | 21, 22 | mpoex 7893 |
. . . . 5
⊢ (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦)) ∈ V |
24 | | eqid 2738 |
. . . . . 6
⊢
({〈(Base‘ndx), 𝐵〉, 〈(+g‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘f
(+g‘𝑀)𝑦))〉, 〈(.r‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘ 𝑦))〉} ∪ {〈(Scalar‘ndx),
𝑆〉, 〈(
·𝑠 ‘ndx), (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦))〉}) = ({〈(Base‘ndx), 𝐵〉,
〈(+g‘ndx), (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘f
(+g‘𝑀)𝑦))〉, 〈(.r‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘ 𝑦))〉} ∪ {〈(Scalar‘ndx),
𝑆〉, 〈(
·𝑠 ‘ndx), (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦))〉}) |
25 | 24 | algvsca 40923 |
. . . . 5
⊢ ((𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦)) ∈ V → (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦)) = ( ·𝑠
‘({〈(Base‘ndx), 𝐵〉, 〈(+g‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘f
(+g‘𝑀)𝑦))〉, 〈(.r‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘ 𝑦))〉} ∪ {〈(Scalar‘ndx),
𝑆〉, 〈(
·𝑠 ‘ndx), (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦))〉}))) |
26 | 23, 25 | mp1i 13 |
. . . 4
⊢ (𝑀 ∈ V → (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦)) = ( ·𝑠
‘({〈(Base‘ndx), 𝐵〉, 〈(+g‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘f
(+g‘𝑀)𝑦))〉, 〈(.r‘ndx),
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐵 ↦ (𝑥 ∘ 𝑦))〉} ∪ {〈(Scalar‘ndx),
𝑆〉, 〈(
·𝑠 ‘ndx), (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦))〉}))) |
27 | 20, 26 | eqtr4d 2781 |
. . 3
⊢ (𝑀 ∈ V → (
·𝑠 ‘(MEndo‘𝑀)) = (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦))) |
28 | | fvprc 6748 |
. . . . . 6
⊢ (¬
𝑀 ∈ V →
(MEndo‘𝑀) =
∅) |
29 | 28 | fveq2d 6760 |
. . . . 5
⊢ (¬
𝑀 ∈ V → (
·𝑠 ‘(MEndo‘𝑀)) = ( ·𝑠
‘∅)) |
30 | | vscaid 16956 |
. . . . . 6
⊢
·𝑠 = Slot (
·𝑠 ‘ndx) |
31 | 30 | str0 16818 |
. . . . 5
⊢ ∅ =
( ·𝑠 ‘∅) |
32 | 29, 31 | eqtr4di 2797 |
. . . 4
⊢ (¬
𝑀 ∈ V → (
·𝑠 ‘(MEndo‘𝑀)) = ∅) |
33 | | fvprc 6748 |
. . . . . . . . 9
⊢ (¬
𝑀 ∈ V →
(Scalar‘𝑀) =
∅) |
34 | 8, 33 | syl5eq 2791 |
. . . . . . . 8
⊢ (¬
𝑀 ∈ V → 𝑆 = ∅) |
35 | 34 | fveq2d 6760 |
. . . . . . 7
⊢ (¬
𝑀 ∈ V →
(Base‘𝑆) =
(Base‘∅)) |
36 | | base0 16845 |
. . . . . . 7
⊢ ∅ =
(Base‘∅) |
37 | 35, 9, 36 | 3eqtr4g 2804 |
. . . . . 6
⊢ (¬
𝑀 ∈ V → 𝐾 = ∅) |
38 | 37 | orcd 869 |
. . . . 5
⊢ (¬
𝑀 ∈ V → (𝐾 = ∅ ∨ 𝐵 = ∅)) |
39 | | 0mpo0 7336 |
. . . . 5
⊢ ((𝐾 = ∅ ∨ 𝐵 = ∅) → (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦)) = ∅) |
40 | 38, 39 | syl 17 |
. . . 4
⊢ (¬
𝑀 ∈ V → (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦)) = ∅) |
41 | 32, 40 | eqtr4d 2781 |
. . 3
⊢ (¬
𝑀 ∈ V → (
·𝑠 ‘(MEndo‘𝑀)) = (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦))) |
42 | 27, 41 | pm2.61i 182 |
. 2
⊢ (
·𝑠 ‘(MEndo‘𝑀)) = (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦)) |
43 | 2, 42 | eqtri 2766 |
1
⊢ (
·𝑠 ‘𝐴) = (𝑥 ∈ 𝐾, 𝑦 ∈ 𝐵 ↦ ((𝐸 × {𝑥}) ∘f · 𝑦)) |