let L be non empty doubleLoopStr ; :: thesis: for a, b being Element of L st 0. L <> 1. L & ( for a being Element of L holds a * (1. L) = a ) & ( for a being Element of L st a <> 0. L holds
ex x being Element of L st a * x = 1. L ) & ( for a, b, c being Element of L holds (a * b) * c = a * (b * c) ) & ( for a being Element of L holds a * (0. L) = 0. L ) & a * b = 1. L holds
b * a = 1. L
let a, b be Element of L; :: thesis: ( 0. L <> 1. L & ( for a being Element of L holds a * (1. L) = a ) & ( for a being Element of L st a <> 0. L holds
ex x being Element of L st a * x = 1. L ) & ( for a, b, c being Element of L holds (a * b) * c = a * (b * c) ) & ( for a being Element of L holds a * (0. L) = 0. L ) & a * b = 1. L implies b * a = 1. L )
assume that
A1: 0. L <> 1. L and
A2: for a being Element of L holds a * (1. L) = a and
A3: for a being Element of L st a <> 0. L holds
ex x being Element of L st a * x = 1. L and
A4: for a, b, c being Element of L holds (a * b) * c = a * (b * c) and
A5: for a being Element of L holds a * (0. L) = 0. L ; :: thesis: ( not a * b = 1. L or b * a = 1. L )
thus ( a * b = 1. L implies b * a = 1. L ) :: thesis: verum

proof

assume A6: a * b = 1. L ; :: thesis: b * a = 1. L
then b <> 0. L by A1, A5;
then consider x being Element of L such that
A7: b * x = 1. L by A3;
thus b * a = (b * a) * (b * x) by A2, A7
.= ((b * a) * b) * x by A4
.= (b * (1. L)) * x by A4, A6
.= 1. L by A2, A7 ; :: thesis: verum

end;