Metamath Proof Explorer


Theorem ressascl

Description: The injection of scalars is invariant between subalgebras and superalgebras. (Contributed by Mario Carneiro, 9-Mar-2015)

Ref Expression
Hypotheses ressascl.a A = algSc W
ressascl.x X = W 𝑠 S
Assertion ressascl S SubRing W A = algSc X

Proof

Step Hyp Ref Expression
1 ressascl.a A = algSc W
2 ressascl.x X = W 𝑠 S
3 eqid Scalar W = Scalar W
4 2 3 resssca S SubRing W Scalar W = Scalar X
5 4 fveq2d S SubRing W Base Scalar W = Base Scalar X
6 eqid W = W
7 2 6 ressvsca S SubRing W W = X
8 eqidd S SubRing W x = x
9 eqid 1 W = 1 W
10 2 9 subrg1 S SubRing W 1 W = 1 X
11 7 8 10 oveq123d S SubRing W x W 1 W = x X 1 X
12 5 11 mpteq12dv S SubRing W x Base Scalar W x W 1 W = x Base Scalar X x X 1 X
13 eqid Base Scalar W = Base Scalar W
14 1 3 13 6 9 asclfval A = x Base Scalar W x W 1 W
15 eqid algSc X = algSc X
16 eqid Scalar X = Scalar X
17 eqid Base Scalar X = Base Scalar X
18 eqid X = X
19 eqid 1 X = 1 X
20 15 16 17 18 19 asclfval algSc X = x Base Scalar X x X 1 X
21 12 14 20 3eqtr4g S SubRing W A = algSc X