Metamath Proof Explorer


Theorem evls1scasrng

Description: The evaluation of a scalar of a subring yields the same result as evaluated as a scalar over the ring itself. (Contributed by AV, 13-Sep-2019)

Ref Expression
Hypotheses evls1scasrng.q Q = S evalSub 1 R
evls1scasrng.o O = eval 1 S
evls1scasrng.w W = Poly 1 U
evls1scasrng.u U = S 𝑠 R
evls1scasrng.p P = Poly 1 S
evls1scasrng.b B = Base S
evls1scasrng.a A = algSc W
evls1scasrng.c C = algSc P
evls1scasrng.s φ S CRing
evls1scasrng.r φ R SubRing S
evls1scasrng.x φ X R
Assertion evls1scasrng φ Q A X = O C X

Proof

Step Hyp Ref Expression
1 evls1scasrng.q Q = S evalSub 1 R
2 evls1scasrng.o O = eval 1 S
3 evls1scasrng.w W = Poly 1 U
4 evls1scasrng.u U = S 𝑠 R
5 evls1scasrng.p P = Poly 1 S
6 evls1scasrng.b B = Base S
7 evls1scasrng.a A = algSc W
8 evls1scasrng.c C = algSc P
9 evls1scasrng.s φ S CRing
10 evls1scasrng.r φ R SubRing S
11 evls1scasrng.x φ X R
12 6 ressid S CRing S 𝑠 B = S
13 12 eqcomd S CRing S = S 𝑠 B
14 9 13 syl φ S = S 𝑠 B
15 14 fveq2d φ Poly 1 S = Poly 1 S 𝑠 B
16 5 15 syl5eq φ P = Poly 1 S 𝑠 B
17 16 fveq2d φ algSc P = algSc Poly 1 S 𝑠 B
18 8 17 syl5eq φ C = algSc Poly 1 S 𝑠 B
19 18 fveq1d φ C X = algSc Poly 1 S 𝑠 B X
20 19 fveq2d φ S evalSub 1 B C X = S evalSub 1 B algSc Poly 1 S 𝑠 B X
21 eqid S evalSub 1 B = S evalSub 1 B
22 eqid Poly 1 S 𝑠 B = Poly 1 S 𝑠 B
23 eqid S 𝑠 B = S 𝑠 B
24 eqid algSc Poly 1 S 𝑠 B = algSc Poly 1 S 𝑠 B
25 crngring S CRing S Ring
26 6 subrgid S Ring B SubRing S
27 9 25 26 3syl φ B SubRing S
28 6 subrgss R SubRing S R B
29 10 28 syl φ R B
30 29 11 sseldd φ X B
31 21 22 23 6 24 9 27 30 evls1sca φ S evalSub 1 B algSc Poly 1 S 𝑠 B X = B × X
32 20 31 eqtrd φ S evalSub 1 B C X = B × X
33 2 6 evl1fval1 O = S evalSub 1 B
34 33 a1i φ O = S evalSub 1 B
35 34 fveq1d φ O C X = S evalSub 1 B C X
36 1 3 4 6 7 9 10 11 evls1sca φ Q A X = B × X
37 32 35 36 3eqtr4rd φ Q A X = O C X