let SAS be Semi_Affine_Space; :: thesis: for a, b, c, d being Element of SAS st parallelogram a,b,c,d holds
not parallelogram a,b,d,c
let a, b, c, d be Element of SAS; :: thesis: ( parallelogram a,b,c,d implies not parallelogram a,b,d,c )
assume A1: parallelogram a,b,c,d ; :: thesis: not parallelogram a,b,d,c
then not a,b,c are_collinear ;
then A2: not a,b // a,c ;
assume parallelogram a,b,d,c ; :: thesis: contradiction
then A3: a,d // b,c ;
( a,b // c,d & a,c // b,d ) by A1;
hence contradiction by A3, A2, Def1; :: thesis: verum