Metamath Proof Explorer


Theorem matbas

Description: The matrix ring has the same base set as its underlying group. (Contributed by Stefan O'Rear, 4-Sep-2015)

Ref Expression
Hypotheses matbas.a ⊢ A = N Mat R
matbas.g ⊢ G = R freeLMod N × N
Assertion matbas ⊢ N ∈ Fin ∧ R ∈ V → Base G = Base A

Proof

Step Hyp Ref Expression
1 matbas.a ⊢ A = N Mat R
2 matbas.g ⊢ G = R freeLMod N × N
3 baseid ⊢ Base = Slot Base ndx
4 basendxnmulrndx ⊢ Base ndx ≠ ⋅ ndx
5 3 4 setsnid ⊢ Base G = Base G sSet ⋅ ndx R maMul N N N
6 eqid ⊢ R maMul N N N = R maMul N N N
7 1 2 6 matval ⊢ N ∈ Fin ∧ R ∈ V → A = G sSet ⋅ ndx R maMul N N N
8 7 fveq2d ⊢ N ∈ Fin ∧ R ∈ V → Base A = Base G sSet ⋅ ndx R maMul N N N
9 5 8 eqtr4id ⊢ N ∈ Fin ∧ R ∈ V → Base G = Base A