let O be Ordinal; for R being non trivial right_complementable associative commutative Abelian add-associative right_zeroed well-unital distributive doubleLoopStr
for t being bag of O
for p being Polynomial of O,R
for i being object
for x being Function of O,R st i in O holds
eval ((Subst (t,i,p)),x) = eval (t,(x +* (i,(eval (p,x)))))
let R be non trivial right_complementable associative commutative Abelian add-associative right_zeroed well-unital distributive doubleLoopStr ; for t being bag of O
for p being Polynomial of O,R
for i being object
for x being Function of O,R st i in O holds
eval ((Subst (t,i,p)),x) = eval (t,(x +* (i,(eval (p,x)))))
let t be bag of O; for p being Polynomial of O,R
for i being object
for x being Function of O,R st i in O holds
eval ((Subst (t,i,p)),x) = eval (t,(x +* (i,(eval (p,x)))))
let p be Polynomial of O,R; for i being object
for x being Function of O,R st i in O holds
eval ((Subst (t,i,p)),x) = eval (t,(x +* (i,(eval (p,x)))))
let i be object ; for x being Function of O,R st i in O holds
eval ((Subst (t,i,p)),x) = eval (t,(x +* (i,(eval (p,x)))))
let x be Function of O,R; ( i in O implies eval ((Subst (t,i,p)),x) = eval (t,(x +* (i,(eval (p,x))))) )
assume A1:
i in O
; eval ((Subst (t,i,p)),x) = eval (t,(x +* (i,(eval (p,x)))))
set xE = x +* (i,(eval (p,x)));
set P = p `^ (t . i);
set t0 = t +* (i,0);
set SS = SgmX ((BagOrder O),(Support (p `^ (t . i))));
set Smp = Subst (t,i,p);
consider y being FinSequence of R such that
A2:
( len y = len (SgmX ((BagOrder O),(Support (p `^ (t . i))))) & eval ((p `^ (t . i)),x) = Sum y )
and
A3:
for i being Element of NAT st 1 <= i & i <= len y holds
y /. i = (((p `^ (t . i)) * (SgmX ((BagOrder O),(Support (p `^ (t . i)))))) /. i) * (eval (((SgmX ((BagOrder O),(Support (p `^ (t . i))))) /. i),x))
by POLYNOM2:def 4;
BagOrder O linearly_orders Support (p `^ (t . i))
by POLYNOM2:18;
then A4:
rng (SgmX ((BagOrder O),(Support (p `^ (t . i))))) = Support (p `^ (t . i))
by PRE_POLY:def 2;
A5:
SgmX ((BagOrder O),(Support (p `^ (t . i)))) is one-to-one
by POLYNOM2:18, PRE_POLY:10;
consider tran being one-to-one FinSequence of Bags O such that
A6:
rng tran = Support (Subst (t,i,p))
and
A7:
len tran = len (SgmX ((BagOrder O),(Support (p `^ (t . i)))))
and
A8:
for j being Nat st 1 <= j & j <= len tran holds
tran . j = Subst (t,i,((SgmX ((BagOrder O),(Support (p `^ (t . i))))) /. j))
by A4, A5, Th34;
A9:
len tran = card (Support (Subst (t,i,p)))
by A6, FINSEQ_4:62;
A10:
card (Support (Subst (t,i,p))) = len (SgmX ((BagOrder O),(Support (p `^ (t . i)))))
by A6, FINSEQ_4:62, A7;
consider Y being FinSequence of R such that
A11:
( len Y = card (Support (Subst (t,i,p))) & eval ((Subst (t,i,p)),x) = Sum Y )
and
A12:
for i being Nat st 1 <= i & i <= len Y holds
Y /. i = (((Subst (t,i,p)) * tran) /. i) * (eval ((tran /. i),x))
by A6, HILB10_2:24;
A13:
dom (y * (eval ((t +* (i,0)),x))) = dom y
by POLYNOM1:def 2;
then A14:
len (y * (eval ((t +* (i,0)),x))) = len y
by FINSEQ_3:29;
A15:
eval ((p `^ (t . i)),x) = (power R) . ((eval (p,x)),(t . i))
by Th30;
for j being Nat st 1 <= j & j <= len Y holds
Y . j = (y * (eval ((t +* (i,0)),x))) . j
proof
let j be
Nat;
( 1 <= j & j <= len Y implies Y . j = (y * (eval ((t +* (i,0)),x))) . j )
assume A16:
( 1
<= j &
j <= len Y )
;
Y . j = (y * (eval ((t +* (i,0)),x))) . j
A17:
j in NAT
by ORDINAL1:def 12;
A18:
(
len Y = len tran &
len tran = len (SgmX ((BagOrder O),(Support (p `^ (t . i))))) &
len (SgmX ((BagOrder O),(Support (p `^ (t . i))))) = len y )
by A2, A11, A6, FINSEQ_4:62, A7;
A19:
rng tran c= Bags O
by RELAT_1:def 19;
(
dom (p `^ (t . i)) = Bags O &
Bags O = dom (Subst (t,i,p)) )
by FUNCT_2:def 1;
then A20:
(
dom ((p `^ (t . i)) * (SgmX ((BagOrder O),(Support (p `^ (t . i)))))) = dom (SgmX ((BagOrder O),(Support (p `^ (t . i))))) &
dom ((Subst (t,i,p)) * tran) = dom tran )
by A19, A4, RELAT_1:27;
A21:
(
j in dom tran &
j in dom (SgmX ((BagOrder O),(Support (p `^ (t . i))))) &
j in dom y &
j in dom Y )
by A16, A2, A11, A7, A9, FINSEQ_3:25;
then A22:
(
tran /. j = tran . j &
tran . j = Subst (
t,
i,
((SgmX ((BagOrder O),(Support (p `^ (t . i))))) /. j)) )
by A16, A8, A11, A9, PARTFUN1:def 6;
A23:
((p `^ (t . i)) * (SgmX ((BagOrder O),(Support (p `^ (t . i)))))) /. j =
((p `^ (t . i)) * (SgmX ((BagOrder O),(Support (p `^ (t . i)))))) . j
by A20, A16, A11, A10, FINSEQ_3:25, PARTFUN1:def 6
.=
(p `^ (t . i)) . ((SgmX ((BagOrder O),(Support (p `^ (t . i))))) . j)
by A20, A16, A11, A10, FINSEQ_3:25, FUNCT_1:12
.=
(p `^ (t . i)) . ((SgmX ((BagOrder O),(Support (p `^ (t . i))))) /. j)
by A21, PARTFUN1:def 6
;
A24:
((Subst (t,i,p)) * tran) /. j =
((Subst (t,i,p)) * tran) . j
by A16, A11, A9, FINSEQ_3:25, A20, PARTFUN1:def 6
.=
(Subst (t,i,p)) . (tran . j)
by A16, A11, A9, FINSEQ_3:25, A20, FUNCT_1:12
.=
((p `^ (t . i)) * (SgmX ((BagOrder O),(Support (p `^ (t . i)))))) /. j
by A23, Th33, A22
;
(
Y . j = Y /. j &
Y /. j = (((Subst (t,i,p)) * tran) /. j) * (eval ((tran /. j),x)) )
by A16, A12, A21, PARTFUN1:def 6;
hence Y . j =
(((p `^ (t . i)) * (SgmX ((BagOrder O),(Support (p `^ (t . i)))))) /. j) * ((eval ((t +* (i,0)),x)) * (eval (((SgmX ((BagOrder O),(Support (p `^ (t . i))))) /. j),x)))
by A22, A24, POLYNOM2:16
.=
((((p `^ (t . i)) * (SgmX ((BagOrder O),(Support (p `^ (t . i)))))) /. j) * (eval (((SgmX ((BagOrder O),(Support (p `^ (t . i))))) /. j),x))) * (eval ((t +* (i,0)),x))
by GROUP_1:def 3
.=
(y /. j) * (eval ((t +* (i,0)),x))
by A17, A3, A18, A16
.=
(y * (eval ((t +* (i,0)),x))) /. j
by A21, POLYNOM1:def 2
.=
(y * (eval ((t +* (i,0)),x))) . j
by A13, A16, A2, A11, A7, A9, FINSEQ_3:25, PARTFUN1:def 6
;
verum
end;
then A25:
Y = y * (eval ((t +* (i,0)),x))
by A14, A2, A11, A6, FINSEQ_4:62, A7, FINSEQ_1:14;
dom t = O
by PARTFUN1:def 2;
then A26:
(t +* (i,0)) . i = 0
by A1, FUNCT_7:31;
( dom x = O & O = dom (x +* (i,(eval (p,x)))) )
by PARTFUN1:def 2;
then A27:
( (x +* (i,(eval (p,x)))) /. i = (x +* (i,(eval (p,x)))) . i & (x +* (i,(eval (p,x)))) . i = eval (p,x) )
by A1, FUNCT_7:31, PARTFUN1:def 6;
( (power R) . (((x +* (i,(eval (p,x)))) /. i),0) = 1_ R & 1_ R = 1. R )
by GROUP_1:def 7;
then
(eval ((t +* (i,0)),(x +* (i,(eval (p,x)))))) * ((power R) . (((x +* (i,(eval (p,x)))) /. i),(t . i))) = (eval (t,(x +* (i,(eval (p,x)))))) * (1. R)
by A1, Th17;
hence eval (t,(x +* (i,(eval (p,x))))) =
((power R) . ((eval (p,x)),(t . i))) * (eval ((t +* (i,0)),x))
by A26, Th18, A27
.=
eval ((Subst (t,i,p)),x)
by A25, A15, A2, A11, POLYNOM1:13
;
verum