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=algScW
ressascl.x X=W𝑠S
Assertion ressascl SSubRingWA=algScX

Proof

Step Hyp Ref Expression
1 ressascl.a A=algScW
2 ressascl.x X=W𝑠S
3 eqid ScalarW=ScalarW
4 2 3 resssca SSubRingWScalarW=ScalarX
5 4 fveq2d SSubRingWBaseScalarW=BaseScalarX
6 eqid W=W
7 2 6 ressvsca SSubRingWW=X
8 eqidd SSubRingWx=x
9 eqid 1W=1W
10 2 9 subrg1 SSubRingW1W=1X
11 7 8 10 oveq123d SSubRingWxW1W=xX1X
12 5 11 mpteq12dv SSubRingWxBaseScalarWxW1W=xBaseScalarXxX1X
13 eqid BaseScalarW=BaseScalarW
14 1 3 13 6 9 asclfval A=xBaseScalarWxW1W
15 eqid algScX=algScX
16 eqid ScalarX=ScalarX
17 eqid BaseScalarX=BaseScalarX
18 eqid X=X
19 eqid 1X=1X
20 15 16 17 18 19 asclfval algScX=xBaseScalarXxX1X
21 12 14 20 3eqtr4g SSubRingWA=algScX