let X be BCI-algebra; :: thesis: for x, y being Element of X st X is alternative & x \ y = 0. X holds
x = y
let x, y be Element of X; :: thesis: ( X is alternative & x \ y = 0. X implies x = y )
assume that
A1: X is alternative and
A2: x \ y = 0. X ; :: thesis: x = y
x \ (x \ y) = x by A2, Th2;
then (x \ x) \ y = x by A1;
then y ` = x by Def5;
hence x = y by A1, Th76; :: thesis: verum