Metamath Proof Explorer


Theorem mplbaspropd

Description: Property deduction for polynomial base set. (Contributed by Stefan O'Rear, 27-Mar-2015) (Proof shortened by AV, 19-Jul-2019)

Ref Expression
Hypotheses psrplusgpropd.b1 φ B = Base R
psrplusgpropd.b2 φ B = Base S
psrplusgpropd.p φ x B y B x + R y = x + S y
Assertion mplbaspropd φ Base I mPoly R = Base I mPoly S

Proof

Step Hyp Ref Expression
1 psrplusgpropd.b1 φ B = Base R
2 psrplusgpropd.b2 φ B = Base S
3 psrplusgpropd.p φ x B y B x + R y = x + S y
4 1 2 eqtr3d φ Base R = Base S
5 4 psrbaspropd φ Base I mPwSer R = Base I mPwSer S
6 5 adantr φ I V Base I mPwSer R = Base I mPwSer S
7 1 2 3 grpidpropd φ 0 R = 0 S
8 7 breq2d φ finSupp 0 R a finSupp 0 S a
9 8 adantr φ I V finSupp 0 R a finSupp 0 S a
10 6 9 rabeqbidv φ I V a Base I mPwSer R | finSupp 0 R a = a Base I mPwSer S | finSupp 0 S a
11 eqid I mPoly R = I mPoly R
12 eqid I mPwSer R = I mPwSer R
13 eqid Base I mPwSer R = Base I mPwSer R
14 eqid 0 R = 0 R
15 eqid Base I mPoly R = Base I mPoly R
16 11 12 13 14 15 mplbas Base I mPoly R = a Base I mPwSer R | finSupp 0 R a
17 eqid I mPoly S = I mPoly S
18 eqid I mPwSer S = I mPwSer S
19 eqid Base I mPwSer S = Base I mPwSer S
20 eqid 0 S = 0 S
21 eqid Base I mPoly S = Base I mPoly S
22 17 18 19 20 21 mplbas Base I mPoly S = a Base I mPwSer S | finSupp 0 S a
23 10 16 22 3eqtr4g φ I V Base I mPoly R = Base I mPoly S
24 reldmmpl Rel dom mPoly
25 24 ovprc1 ¬ I V I mPoly R =
26 24 ovprc1 ¬ I V I mPoly S =
27 25 26 eqtr4d ¬ I V I mPoly R = I mPoly S
28 27 fveq2d ¬ I V Base I mPoly R = Base I mPoly S
29 28 adantl φ ¬ I V Base I mPoly R = Base I mPoly S
30 23 29 pm2.61dan φ Base I mPoly R = Base I mPoly S