theorem Th21:
  n = 1 & m = 1 implies Segm(A,nt,mt) = <*<* A*(nt.1,mt.1) *>*>
proof
A1: 1 in Seg 1;
  assume that
A2: n = 1 and
A3: m = 1;
  Indices Segm(A,nt,mt)=[:Seg 1,Seg 1:] by A2,A3,MATRIX_0:24;
  then [1,1] in Indices Segm(A,nt,mt) by A1,ZFMISC_1:87;
  then Segm(A,nt,mt)*(1,1)=A*(nt.1,mt.1) by Def1;
  hence thesis by A2,A3,Th20;
end;
