let M be MidSp; :: thesis: ( ( for a being Element of (vectgroup M) ex x being Element of (vectgroup M) st Double x = a ) & ( for a being Element of (vectgroup M) st Double a = 0. (vectgroup M) holds
a = 0. (vectgroup M) ) )
set G = vectgroup M;
set p = the Point of M;
thus for a being Element of (vectgroup M) ex x being Element of (vectgroup M) st Double x = a :: thesis: for a being Element of (vectgroup M) st Double a = 0. (vectgroup M) holds
a = 0. (vectgroup M)
proof
set p = the Point of M;
let a be Element of (vectgroup M); :: thesis: ex x being Element of (vectgroup M) st Double x = a
reconsider aa = a as Vector of M by Th15;
consider q being Point of M such that
A1: aa = vect ( the Point of M,q) by MIDSP_1:35;
set xx = vect ( the Point of M,( the Point of M @ q));
reconsider x = vect ( the Point of M,( the Point of M @ q)) as Element of (vectgroup M) by Th15;
take x ; :: thesis: Double x = a
(vect ( the Point of M,( the Point of M @ q))) + (vect ( the Point of M,( the Point of M @ q))) = aa by A1, MIDSP_1:42;
hence Double x = a by Th15; :: thesis: verum
end;
let a be Element of (vectgroup M); :: thesis: ( Double a = 0. (vectgroup M) implies a = 0. (vectgroup M) )
reconsider aa = a as Vector of M by Th15;
consider q being Point of M such that
A2: aa = vect ( the Point of M,q) by MIDSP_1:35;
consider r being Point of M such that
A3: aa = vect (q,r) by MIDSP_1:35;
assume Double a = 0. (vectgroup M) ; :: thesis: a = 0. (vectgroup M)
then aa + aa = 0. (vectgroup M) by Th15
.= ID M by Th15 ;
then (vect ( the Point of M,q)) + (vect (q,r)) = vect ( the Point of M, the Point of M) by A2, A3, MIDSP_1:38;
then vect ( the Point of M,r) = vect ( the Point of M, the Point of M) by MIDSP_1:40;
then vect ( the Point of M,q) = vect (q, the Point of M) by A2, A3, MIDSP_1:39;
then A4: the Point of M @ the Point of M = q @ q by MIDSP_1:37;
the Point of M = the Point of M @ the Point of M by MIDSP_1:def 3
.= q by A4, MIDSP_1:def 3 ;
hence a = ID M by A2, MIDSP_1:38
.= 0. (vectgroup M) by Th15 ;
:: thesis: verum