theorem Th23:
  n = 2 & m = 2 implies Segm(A,nt,mt) = ( A*(nt.1,mt.1),A*(nt.1,mt
  .2) )][( A*(nt.2,mt.1),A*(nt.2,mt.2) )
proof
  set S=Segm(A,nt,mt);
  set I=Indices S;
  assume that
A1: n = 2 and
A2: m = 2;
A3: I=[:Seg 2,Seg 2:] by A1,A2,MATRIX_0:24;
A4: 2 in Seg 2;
  then [2,2] in I by A3,ZFMISC_1:87;
  then
A5: S*(2,2)=A*(nt.2,mt.2) by Def1;
A6: 1 in Seg 2;
  then [1,1] in I by A3,ZFMISC_1:87;
  then
A7: S*(1,1)=A*(nt.1,mt.1) by Def1;
  [2,1] in I by A6,A4,A3,ZFMISC_1:87;
  then
A8: S*(2,1)=A*(nt.2,mt.1) by Def1;
  [1,2] in I by A6,A4,A3,ZFMISC_1:87;
  then S*(1,2)=A*(nt.1,mt.2) by Def1;
  hence thesis by A1,A2,A7,A8,A5,Th22;
end;
