Metamath Proof Explorer


Theorem ply1coe1eq

Description: Two polynomials over the same ring are equal iff they have identical coefficients. (Contributed by AV, 13-Oct-2019)

Ref Expression
Hypotheses eqcoe1ply1eq.p P = Poly 1 R
eqcoe1ply1eq.b B = Base P
eqcoe1ply1eq.a A = coe 1 K
eqcoe1ply1eq.c C = coe 1 L
Assertion ply1coe1eq R Ring K B L B k 0 A k = C k K = L

Proof

Step Hyp Ref Expression
1 eqcoe1ply1eq.p P = Poly 1 R
2 eqcoe1ply1eq.b B = Base P
3 eqcoe1ply1eq.a A = coe 1 K
4 eqcoe1ply1eq.c C = coe 1 L
5 1 2 3 4 eqcoe1ply1eq R Ring K B L B k 0 A k = C k K = L
6 fveq2 K = L coe 1 K = coe 1 L
7 6 adantl R Ring K B L B K = L coe 1 K = coe 1 L
8 7 3 4 3eqtr4g R Ring K B L B K = L A = C
9 8 adantr R Ring K B L B K = L k 0 A = C
10 9 fveq1d R Ring K B L B K = L k 0 A k = C k
11 10 ralrimiva R Ring K B L B K = L k 0 A k = C k
12 11 ex R Ring K B L B K = L k 0 A k = C k
13 5 12 impbid R Ring K B L B k 0 A k = C k K = L