Metamath Proof Explorer


Theorem ply1scl0

Description: The zero scalar is zero. (Contributed by Stefan O'Rear, 29-Mar-2015)

Ref Expression
Hypotheses ply1scl.p P = Poly 1 R
ply1scl.a A = algSc P
ply1scl0.z 0 ˙ = 0 R
ply1scl0.y Y = 0 P
Assertion ply1scl0 R Ring A 0 ˙ = Y

Proof

Step Hyp Ref Expression
1 ply1scl.p P = Poly 1 R
2 ply1scl.a A = algSc P
3 ply1scl0.z 0 ˙ = 0 R
4 ply1scl0.y Y = 0 P
5 eqid Base R = Base R
6 5 3 ring0cl R Ring 0 ˙ Base R
7 1 ply1sca2 I R = Scalar P
8 df-base Base = Slot 1
9 8 5 strfvi Base R = Base I R
10 eqid P = P
11 eqid 1 P = 1 P
12 2 7 9 10 11 asclval 0 ˙ Base R A 0 ˙ = 0 ˙ P 1 P
13 6 12 syl R Ring A 0 ˙ = 0 ˙ P 1 P
14 fvi R Ring I R = R
15 14 fveq2d R Ring 0 I R = 0 R
16 3 15 eqtr4id R Ring 0 ˙ = 0 I R
17 16 oveq1d R Ring 0 ˙ P 1 P = 0 I R P 1 P
18 1 ply1lmod R Ring P LMod
19 1 ply1ring R Ring P Ring
20 eqid Base P = Base P
21 20 11 ringidcl P Ring 1 P Base P
22 19 21 syl R Ring 1 P Base P
23 eqid 0 I R = 0 I R
24 20 7 10 23 4 lmod0vs P LMod 1 P Base P 0 I R P 1 P = Y
25 18 22 24 syl2anc R Ring 0 I R P 1 P = Y
26 13 17 25 3eqtrd R Ring A 0 ˙ = Y