let A be non empty set ; :: thesis: for x, y, z being Element of (CAlgebra A)
for a, b being Complex holds
( x + y = y + x & (x + y) + z = x + (y + z) & x + (0. (CAlgebra A)) = x & x is right_complementable & x * y = y * x & (x * y) * z = x * (y * z) & x * (1. (CAlgebra A)) = x & x * (y + z) = (x * y) + (x * z) & a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
let x, y, z be Element of (CAlgebra A); :: thesis: for a, b being Complex holds
( x + y = y + x & (x + y) + z = x + (y + z) & x + (0. (CAlgebra A)) = x & x is right_complementable & x * y = y * x & (x * y) * z = x * (y * z) & x * (1. (CAlgebra A)) = x & x * (y + z) = (x * y) + (x * z) & a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
let a, b be Complex; :: thesis: ( x + y = y + x & (x + y) + z = x + (y + z) & x + (0. (CAlgebra A)) = x & x is right_complementable & x * y = y * x & (x * y) * z = x * (y * z) & x * (1. (CAlgebra A)) = x & x * (y + z) = (x * y) + (x * z) & a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
set IT = CAlgebra A;
reconsider f = x as Element of Funcs A,COMPLEX ;
thus x + y = y + x by Th7; :: thesis: ( (x + y) + z = x + (y + z) & x + (0. (CAlgebra A)) = x & x is right_complementable & x * y = y * x & (x * y) * z = x * (y * z) & x * (1. (CAlgebra A)) = x & x * (y + z) = (x * y) + (x * z) & a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
thus (x + y) + z = x + (y + z) by Th8; :: thesis: ( x + (0. (CAlgebra A)) = x & x is right_complementable & x * y = y * x & (x * y) * z = x * (y * z) & x * (1. (CAlgebra A)) = x & x * (y + z) = (x * y) + (x * z) & a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
thus x + (0. (CAlgebra A)) = (ComplexFuncAdd A) . (ComplexFuncZero A),f by Th7
.= x by Th12 ; :: thesis: ( x is right_complementable & x * y = y * x & (x * y) * z = x * (y * z) & x * (1. (CAlgebra A)) = x & x * (y + z) = (x * y) + (x * z) & a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
thus ex t being Element of (CAlgebra A) st x + t = 0. (CAlgebra A) :: according to ALGSTR_0:def 11 :: thesis: ( x * y = y * x & (x * y) * z = x * (y * z) & x * (1. (CAlgebra A)) = x & x * (y + z) = (x * y) + (x * z) & a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )

proof

set h = (ComplexFuncExtMult A) . [(- 1r ),f];
reconsider t = (ComplexFuncExtMult A) . [(- 1r ),f] as Element of (CAlgebra A) ;
take t ; :: thesis: x + t = 0. (CAlgebra A)
thus x + t = 0. (CAlgebra A) by Th13; :: thesis: verum

end;

thus x * y = y * x by Th9; :: thesis: ( (x * y) * z = x * (y * z) & x * (1. (CAlgebra A)) = x & x * (y + z) = (x * y) + (x * z) & a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
thus (x * y) * z = x * (y * z) by Th10; :: thesis: ( x * (1. (CAlgebra A)) = x & x * (y + z) = (x * y) + (x * z) & a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
thus x * (1. (CAlgebra A)) = (ComplexFuncMult A) . (ComplexFuncUnit A),f by Th9
.= x by Th11 ; :: thesis: ( x * (y + z) = (x * y) + (x * z) & a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
thus x * (y + z) = (x * y) + (x * z) by Th17; :: thesis: ( a * (x * y) = (a * x) * y & a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
thus a * (x * y) = (a * x) * y by Th18; :: thesis: ( a * (x + y) = (a * x) + (a * y) & (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
thus a * (x + y) = (a * x) + (a * y) by Lm2; :: thesis: ( (a + b) * x = (a * x) + (b * x) & (a * b) * x = a * (b * x) )
thus (a + b) * x = (a * x) + (b * x) by Th16; :: thesis: (a * b) * x = a * (b * x)
thus (a * b) * x = a * (b * x) by Th15; :: thesis: verum