let L be non empty doubleLoopStr ; :: thesis: ( L is _Skew-Field iff ( ( for a being Element of L holds a + (0. L) = a ) & ( for a being Element of L ex x being Element of L st a + x = 0. L ) & ( for a, b, c being Element of L holds (a + b) + c = a + (b + c) ) & ( for a, b being Element of L holds a + b = b + a ) & 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 being Element of L holds a * (0. L) = 0. L ) & ( for a being Element of L holds (0. L) * a = 0. L ) & ( for a, b, c being Element of L holds (a * b) * c = a * (b * c) ) & ( for a, b, c being Element of L holds a * (b + c) = (a * b) + (a * c) ) & ( for a, b, c being Element of L holds (b + c) * a = (b * a) + (c * a) ) ) )

thus ( L is _Skew-Field implies ( ( for a being Element of L holds a + (0. L) = a ) & ( for a being Element of L ex x being Element of L st a + x = 0. L ) & ( for a, b, c being Element of L holds (a + b) + c = a + (b + c) ) & ( for a, b being Element of L holds a + b = b + a ) & 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 being Element of L holds a * (0. L) = 0. L ) & ( for a being Element of L holds (0. L) * a = 0. L ) & ( for a, b, c being Element of L holds (a * b) * c = a * (b * c) ) & ( for a, b, c being Element of L holds a * (b + c) = (a * b) + (a * c) ) & ( for a, b, c being Element of L holds (b + c) * a = (b * a) + (c * a) ) ) ) by ; :: thesis: ( ( for a being Element of L holds a + (0. L) = a ) & ( for a being Element of L ex x being Element of L st a + x = 0. L ) & ( for a, b, c being Element of L holds (a + b) + c = a + (b + c) ) & ( for a, b being Element of L holds a + b = b + a ) & 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 being Element of L holds a * (0. L) = 0. L ) & ( for a being Element of L holds (0. L) * a = 0. L ) & ( for a, b, c being Element of L holds (a * b) * c = a * (b * c) ) & ( for a, b, c being Element of L holds a * (b + c) = (a * b) + (a * c) ) & ( for a, b, c being Element of L holds (b + c) * a = (b * a) + (c * a) ) implies L is _Skew-Field )

assume A1: ( ( for a being Element of L holds a + (0. L) = a ) & ( for a being Element of L ex x being Element of L st a + x = 0. L ) & ( for a, b, c being Element of L holds (a + b) + c = a + (b + c) ) & ( for a, b being Element of L holds a + b = b + a ) & 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 being Element of L holds a * (0. L) = 0. L ) & ( for a being Element of L holds (0. L) * a = 0. L ) & ( for a, b, c being Element of L holds (a * b) * c = a * (b * c) ) & ( for a, b, c being Element of L holds a * (b + c) = (a * b) + (a * c) ) & ( for a, b, c being Element of L holds (b + c) * a = (b * a) + (c * a) ) ) ; :: thesis: L is _Skew-Field
now :: thesis: ( ( for a being Element of L holds (0. L) + a = a ) & ( for a, b being Element of L ex x being Element of L st a + x = b ) & ( for a, b being Element of L ex x being Element of L st x + a = b ) & ( for a, x, y being Element of L st a + x = a + y holds
x = y ) & ( for a, x, y being Element of L st x + a = y + a holds
x = y ) & ( for a being Element of L holds (1. L) * a = a ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st a * x = b ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st x * a = b ) & ( for a, x, y being Element of L st a <> 0. L & a * x = a * y holds
x = y ) & ( for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y ) )
thus A2: for a being Element of L holds (0. L) + a = a :: thesis: ( ( for a, b being Element of L ex x being Element of L st a + x = b ) & ( for a, b being Element of L ex x being Element of L st x + a = b ) & ( for a, x, y being Element of L st a + x = a + y holds
x = y ) & ( for a, x, y being Element of L st x + a = y + a holds
x = y ) & ( for a being Element of L holds (1. L) * a = a ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st a * x = b ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st x * a = b ) & ( for a, x, y being Element of L st a <> 0. L & a * x = a * y holds
x = y ) & ( for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y ) )
proof
let a be Element of L; :: thesis: (0. L) + a = a
thus (0. L) + a = a + (0. L) by A1
.= a by A1 ; :: thesis: verum
end;
thus for a, b being Element of L ex x being Element of L st a + x = b :: thesis: ( ( for a, b being Element of L ex x being Element of L st x + a = b ) & ( for a, x, y being Element of L st a + x = a + y holds
x = y ) & ( for a, x, y being Element of L st x + a = y + a holds
x = y ) & ( for a being Element of L holds (1. L) * a = a ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st a * x = b ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st x * a = b ) & ( for a, x, y being Element of L st a <> 0. L & a * x = a * y holds
x = y ) & ( for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y ) )
proof
let a, b be Element of L; :: thesis: ex x being Element of L st a + x = b
consider y being Element of L such that
A3: a + y = 0. L by A1;
take x = y + b; :: thesis: a + x = b
thus a + x = (0. L) + b by A1, A3
.= b by A2 ; :: thesis: verum
end;
thus for a, b being Element of L ex x being Element of L st x + a = b :: thesis: ( ( for a, x, y being Element of L st a + x = a + y holds
x = y ) & ( for a, x, y being Element of L st x + a = y + a holds
x = y ) & ( for a being Element of L holds (1. L) * a = a ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st a * x = b ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st x * a = b ) & ( for a, x, y being Element of L st a <> 0. L & a * x = a * y holds
x = y ) & ( for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y ) )
proof
let a, b be Element of L; :: thesis: ex x being Element of L st x + a = b
consider y being Element of L such that
A4: y + a = 0. L by ;
take x = b + y; :: thesis: x + a = b
thus x + a = b + (0. L) by A1, A4
.= b by A1 ; :: thesis: verum
end;
thus for a, x, y being Element of L st a + x = a + y holds
x = y :: thesis: ( ( for a, x, y being Element of L st x + a = y + a holds
x = y ) & ( for a being Element of L holds (1. L) * a = a ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st a * x = b ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st x * a = b ) & ( for a, x, y being Element of L st a <> 0. L & a * x = a * y holds
x = y ) & ( for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y ) )
proof
let a, x, y be Element of L; :: thesis: ( a + x = a + y implies x = y )
consider z being Element of L such that
A5: z + a = 0. L by ;
assume a + x = a + y ; :: thesis: x = y
then (z + a) + x = z + (a + y) by A1
.= (z + a) + y by A1 ;
hence x = (0. L) + y by A2, A5
.= y by A2 ;
:: thesis: verum
end;
thus for a, x, y being Element of L st x + a = y + a holds
x = y :: thesis: ( ( for a being Element of L holds (1. L) * a = a ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st a * x = b ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st x * a = b ) & ( for a, x, y being Element of L st a <> 0. L & a * x = a * y holds
x = y ) & ( for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y ) )
proof
let a, x, y be Element of L; :: thesis: ( x + a = y + a implies x = y )
consider z being Element of L such that
A6: a + z = 0. L by A1;
assume x + a = y + a ; :: thesis: x = y
then x + (a + z) = (y + a) + z by A1
.= y + (a + z) by A1 ;
hence x = y + (0. L) by A1, A6
.= y by A1 ;
:: thesis: verum
end;
thus A7: for a being Element of L holds (1. L) * a = a :: thesis: ( ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st a * x = b ) & ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st x * a = b ) & ( for a, x, y being Element of L st a <> 0. L & a * x = a * y holds
x = y ) & ( for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y ) )
proof
let a be Element of L; :: thesis: (1. L) * a = a
thus (1. L) * a = a * (1. L) by
.= a by A1 ; :: thesis: verum
end;
thus for a, b being Element of L st a <> 0. L holds
ex x being Element of L st a * x = b :: thesis: ( ( for a, b being Element of L st a <> 0. L holds
ex x being Element of L st x * a = b ) & ( for a, x, y being Element of L st a <> 0. L & a * x = a * y holds
x = y ) & ( for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y ) )
proof
let a, b be Element of L; :: thesis: ( a <> 0. L implies ex x being Element of L st a * x = b )
assume a <> 0. L ; :: thesis: ex x being Element of L st a * x = b
then consider y being Element of L such that
A8: a * y = 1. L by A1;
take x = y * b; :: thesis: a * x = b
thus a * x = (1. L) * b by A1, A8
.= b by A7 ; :: thesis: verum
end;
thus for a, b being Element of L st a <> 0. L holds
ex x being Element of L st x * a = b :: thesis: ( ( for a, x, y being Element of L st a <> 0. L & a * x = a * y holds
x = y ) & ( for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y ) )
proof
let a, b be Element of L; :: thesis: ( a <> 0. L implies ex x being Element of L st x * a = b )
assume a <> 0. L ; :: thesis: ex x being Element of L st x * a = b
then consider y being Element of L such that
A9: y * a = 1. L by ;
take x = b * y; :: thesis: x * a = b
thus x * a = b * (1. L) by A1, A9
.= b by A1 ; :: thesis: verum
end;
thus for a, x, y being Element of L st a <> 0. L & a * x = a * y holds
x = y :: thesis: for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y
proof
let a, x, y be Element of L; :: thesis: ( a <> 0. L & a * x = a * y implies x = y )
assume a <> 0. L ; :: thesis: ( not a * x = a * y or x = y )
then consider z being Element of L such that
A10: z * a = 1. L by ;
assume a * x = a * y ; :: thesis: x = y
then (z * a) * x = z * (a * y) by A1
.= (z * a) * y by A1 ;
hence x = (1. L) * y by
.= y by A7 ;
:: thesis: verum
end;
thus for a, x, y being Element of L st a <> 0. L & x * a = y * a holds
x = y :: thesis: verum
proof
let a, x, y be Element of L; :: thesis: ( a <> 0. L & x * a = y * a implies x = y )
assume a <> 0. L ; :: thesis: ( not x * a = y * a or x = y )
then consider z being Element of L such that
A11: a * z = 1. L by A1;
assume x * a = y * a ; :: thesis: x = y
then x * (a * z) = (y * a) * z by A1
.= y * (a * z) by A1 ;
hence x = y * (1. L) by
.= y by A1 ;
:: thesis: verum
end;
end;
hence L is _Skew-Field by ; :: thesis: verum