A1: 0_ (n,L) in { x where x is Series of n,L : verum } ;
then reconsider Ca = { x where x is Series of n,L : verum } as non empty set ;
reconsider Ze = 0_ (n,L) as Element of Ca by A1;
defpred S₁[ set , set , set ] means ex p, q being Series of n,L st
( p = $1 & q = $2 & $3 = p + q );
A2: for x, y being Element of Ca ex u being Element of Ca st S₁[x,y,u]

let x, y be Element of Ca; :: thesis:
x in Ca ;
then consider p being Series of n,L such that
A3: x = p ;
y in Ca ;
then consider q being Series of n,L such that
A4: y = q ;
p + q in Ca ;
then reconsider u = p + q as Element of Ca ;
take u ; :: thesis:
take p ; :: thesis:
take q ; :: thesis:
thus ( p = x & q = y & u = p + q ) by A3, A4; :: thesis:

end;

consider Ad being Function of [:Ca,Ca:],Ca such that
A5: for x, y being Element of Ca holds S₁[x,y,Ad . (x,y)] from BINOP_1:sch 3(A2);
1_ (n,L) in { x where x is Series of n,L : verum } ;
then reconsider Un = 1_ (n,L) as Element of Ca ;
defpred S₂[ set , set , set ] means ex p, q being Series of n,L st
( p = $1 & q = $2 & $3 = p *' q );
A6: for x, y being Element of Ca ex u being Element of Ca st S₂[x,y,u]

proof

let x, y be Element of Ca; :: thesis:
x in Ca ;
then consider p being Series of n,L such that
A7: x = p ;
y in Ca ;
then consider q being Series of n,L such that
A8: y = q ;
p *' q in Ca ;
then reconsider u = p *' q as Element of Ca ;
take u ; :: thesis:
take p ; :: thesis:
take q ; :: thesis:
thus ( p = x & q = y & u = p *' q ) by A7, A8; :: thesis:

end;

consider Mu being Function of [:Ca,Ca:],Ca such that
A9: for x, y being Element of Ca holds S₂[x,y,Mu . (x,y)] from BINOP_1:sch 3(A6);
defpred S₃[ Element of L, set , set ] means ex p being Series of n,L st
( p = $2 & $3 = $1 * p );
A10: for a being Element of L
for x being Element of Ca ex u being Element of Ca st S₃[a,x,u]

proof

let a be Element of L; :: thesis:
let x be Element of Ca; :: thesis:
x in Ca ;
then consider q being Series of n,L such that
A11: x = q ;
a * q in Ca ;
then reconsider u = a * q as Element of Ca ;
take u ; :: thesis:
take q ; :: thesis:
thus ( q = x & u = a * q ) by A11; :: thesis:

end;

consider lm being Function of [: the carrier of L,Ca:],Ca such that
A12: for a being Element of L
for x being Element of Ca holds S₃[a,x,lm . (a,x)] from BINOP_1:sch 3(A10);
reconsider P = AlgebraStr(# Ca,Ad,Mu,Ze,Un,lm #) as non empty strict AlgebraStr over L ;
take P ; :: thesis:
thus for x being set holds
( x in the carrier of P iff x is Series of n,L ) :: thesis:

proof

let x be set ; :: thesis:
thus ( x in the carrier of P implies x is Series of n,L ) :: thesis:

proof

assume x in the carrier of P ; :: thesis:
then ex p being Series of n,L st x = p ;
hence x is Series of n,L ; :: thesis:

end;

thus ( x is Series of n,L implies x in the carrier of P ) ; :: thesis:

end;

thus for x, y being Element of P
for p, q being Series of n,L st x = p & y = q holds
x + y = p + q :: thesis:

proof

let x, y be Element of P; :: thesis:
let p, q be Series of n,L; :: thesis:
assume A13: ( x = p & y = q ) ; :: thesis:
ex p1, q1 being Series of n,L st
( p1 = x & q1 = y & Ad . (x,y) = p1 + q1 ) by A5;
hence x + y = p + q by A13; :: thesis:

end;

thus for x, y being Element of P
for p, q being Series of n,L st x = p & y = q holds
x * y = p *' q :: thesis:

proof

let x, y be Element of P; :: thesis:
let p, q be Series of n,L; :: thesis:
assume A14: ( x = p & y = q ) ; :: thesis:
ex p1, q1 being Series of n,L st
( p1 = x & q1 = y & Mu . (x,y) = p1 *' q1 ) by A9;
hence x * y = p *' q by A14; :: thesis:

end;

thus for a being Element of L
for x being Element of P
for p being Series of n,L st x = p holds
a * x = a * p :: thesis:

proof

let a be Element of L; :: thesis:
let x be Element of P; :: thesis:
let p be Series of n,L; :: thesis:
assume A15: x = p ; :: thesis:
ex p1 being Series of n,L st
( x = p1 & lm . (a,x) = a * p1 ) by A12;
hence a * x = a * p by A15; :: thesis:

end;

thus 0. P = 0_ (n,L) ; :: thesis:
thus 1. P = 1_ (n,L) ; :: thesis: