reserve m,n for Element of NAT,
  i,j for Instruction of SCMPDS,
  I for Program
  of SCMPDS,
  a for Int_position;

theorem Th1:
  (i ';' j ';' I). 0=i & (i ';' j ';' I). 1=j
proof
  set jI=j ';' I;
A1: i ';' j ';' I =i ';' jI by SCMPDS_4:16
    .=Load i ';' jI by SCMPDS_4:def 2;
   0 in dom Load i by COMPOS_1:50;
  hence (i ';' j ';' I). 0 =(Load i). 0 by A1,AFINSQ_1:def 3
    .=i;
A2:  0 in dom Load j by COMPOS_1:50;
  0 < card jI;
  then
A3: card Load i=1 &  0 in dom jI by AFINSQ_1:66,COMPOS_1:54;
  thus (i ';' j ';' I). 1 =(Load i ';' jI). (0+1) by A1
    .=jI. 0 by A3,AFINSQ_1:def 3
    .=(Load j ';' I). 0 by SCMPDS_4:def 2
    .=(Load j). 0 by A2,AFINSQ_1:def 3
    .=j;
end;
