let X be BCI-algebra; :: thesis: for x, a, b being Element of X st X is alternative & x \ a = x \ b holds
a = b
let x, a, b be Element of X; :: thesis: ( X is alternative & x \ a = x \ b implies a = b )
assume that
A1: X is alternative and
A2: x \ a = x \ b ; :: thesis: a = b
(x \ x) \ a = x \ (x \ b) by A1, A2;
then (x \ x) \ a = (x \ x) \ b by A1;
then a ` = (x \ x) \ b by Def5;
then a ` = b ` by Def5;
then a = b ` by A1, Th76;
hence a = b by A1, Th76; :: thesis: verum