Metamath Proof Explorer


Theorem fprodconst

Description: The product of constant terms ( k is not free in B ). (Contributed by Scott Fenton, 12-Jan-2018)

Ref Expression
Assertion fprodconst AFinBkAB=BA

Proof

Step Hyp Ref Expression
1 exp0 BB0=1
2 1 eqcomd B1=B0
3 prodeq1 A=kAB=kB
4 prod0 kB=1
5 3 4 eqtrdi A=kAB=1
6 fveq2 A=A=
7 hash0 =0
8 6 7 eqtrdi A=A=0
9 8 oveq2d A=BA=B0
10 5 9 eqeq12d A=kAB=BA1=B0
11 2 10 syl5ibrcom BA=kAB=BA
12 11 adantl AFinBA=kAB=BA
13 eqidd k=fnB=B
14 simprl AFinBAf:1A1-1 ontoAA
15 simprr AFinBAf:1A1-1 ontoAf:1A1-1 ontoA
16 simpllr AFinBAf:1A1-1 ontoAkAB
17 simpllr AFinBAf:1A1-1 ontoAn1AB
18 elfznn n1An
19 18 adantl AFinBAf:1A1-1 ontoAn1An
20 fvconst2g Bn×Bn=B
21 17 19 20 syl2anc AFinBAf:1A1-1 ontoAn1A×Bn=B
22 13 14 15 16 21 fprod AFinBAf:1A1-1 ontoAkAB=seq1××BA
23 expnnval BABA=seq1××BA
24 23 ad2ant2lr AFinBAf:1A1-1 ontoABA=seq1××BA
25 22 24 eqtr4d AFinBAf:1A1-1 ontoAkAB=BA
26 25 expr AFinBAf:1A1-1 ontoAkAB=BA
27 26 exlimdv AFinBAff:1A1-1 ontoAkAB=BA
28 27 expimpd AFinBAff:1A1-1 ontoAkAB=BA
29 fz1f1o AFinA=Aff:1A1-1 ontoA
30 29 adantr AFinBA=Aff:1A1-1 ontoA
31 12 28 30 mpjaod AFinBkAB=BA