A1: 0_. L in { x where x is Polynomial of L : verum } ;
then reconsider Ca = { x where x is Polynomial of L : verum } as non empty set ;
reconsider Ze = 0_. L as Element of Ca by A1;
defpred S1[ set , set , set ] means ex p, q being Polynomial of 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]
proof
let x, y be Element of Ca; :: thesis: ex u being Element of Ca st S1[x,y,u]
x in Ca ;
then consider p being Polynomial of L such that
A3: x = p ;
y in Ca ;
then consider q being Polynomial of L such that
A4: y = q ;
p + q in Ca ;
then reconsider u = p + q as Element of Ca ;
take u ; :: thesis: S1[x,y,u]
take p ; :: thesis: ex q being Polynomial of L st
( p = x & q = y & u = p + q )

take q ; :: thesis: ( p = x & q = y & u = p + q )
thus ( p = x & q = y & u = p + q ) by A3, A4; :: thesis: verum
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 1_. L in { x where x is Polynomial of L : verum } ;
then reconsider Un = 1_. L as Element of Ca ;
defpred S2[ set , set , set ] means ex p, q being Polynomial of 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: ex u being Element of Ca st S2[x,y,u]
x in Ca ;
then consider p being Polynomial of L such that
A7: x = p ;
y in Ca ;
then consider q being Polynomial of L such that
A8: y = q ;
p *' q in Ca ;
then reconsider u = p *' q as Element of Ca ;
take u ; :: thesis: S2[x,y,u]
take p ; :: thesis: ex q being Polynomial of L st
( p = x & q = y & u = p *' q )

take q ; :: thesis: ( p = x & q = y & u = p *' q )
thus ( p = x & q = y & u = p *' q ) by A7, A8; :: thesis: verum
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 reconsider P = doubleLoopStr(# Ca,Ad,Mu,Un,Ze #) as non empty strict doubleLoopStr ;
take P ; :: thesis: ( ( for x being set holds
( x in the carrier of P iff x is Polynomial of L ) ) & ( for x, y being Element of P
for p, q being sequence of L st x = p & y = q holds
x + y = p + q ) & ( for x, y being Element of P
for p, q being sequence of L st x = p & y = q holds
x * y = p *' q ) & 0. P = 0_. L & 1. P = 1_. L )

thus for x being set holds
( x in the carrier of P iff x is Polynomial of L ) :: thesis: ( ( for x, y being Element of P
for p, q being sequence of L st x = p & y = q holds
x + y = p + q ) & ( for x, y being Element of P
for p, q being sequence of L st x = p & y = q holds
x * y = p *' q ) & 0. P = 0_. L & 1. P = 1_. L )
proof
let x be set ; :: thesis: ( x in the carrier of P iff x is Polynomial of L )
thus ( x in the carrier of P implies x is Polynomial of L ) :: thesis: ( x is Polynomial of L implies x in the carrier of P )
proof
assume x in the carrier of P ; :: thesis: x is Polynomial of L
then ex p being Polynomial of L st x = p ;
hence x is Polynomial of L ; :: thesis: verum
end;
thus ( x is Polynomial of L implies x in the carrier of P ) ; :: thesis: verum
end;
thus for x, y being Element of P
for p, q being sequence of L st x = p & y = q holds
x + y = p + q :: thesis: ( ( for x, y being Element of P
for p, q being sequence of L st x = p & y = q holds
x * y = p *' q ) & 0. P = 0_. L & 1. P = 1_. L )
proof
let x, y be Element of P; :: thesis: for p, q being sequence of L st x = p & y = q holds
x + y = p + q

let p, q be sequence of L; :: thesis: ( x = p & y = q implies x + y = p + q )
assume A10: ( x = p & y = q ) ; :: thesis: x + y = p + q
ex p1, q1 being Polynomial of L st
( p1 = x & q1 = y & Ad . (x,y) = p1 + q1 ) by A5;
hence x + y = p + q by A10; :: thesis: verum
end;
thus for x, y being Element of P
for p, q being sequence of L st x = p & y = q holds
x * y = p *' q :: thesis: ( 0. P = 0_. L & 1. P = 1_. L )
proof
let x, y be Element of P; :: thesis: for p, q being sequence of L st x = p & y = q holds
x * y = p *' q

let p, q be sequence of L; :: thesis: ( x = p & y = q implies x * y = p *' q )
assume A11: ( x = p & y = q ) ; :: thesis: x * y = p *' q
ex p1, q1 being Polynomial of L st
( p1 = x & q1 = y & Mu . (x,y) = p1 *' q1 ) by A9;
hence x * y = p *' q by A11; :: thesis: verum
end;
thus 0. P = 0_. L ; :: thesis: 1. P = 1_. L
thus 1. P = 1_. L ; :: thesis: verum