theorem Th45:
  i0 < j0 & n0 < m0 implies Segm(A,{i0,j0},{n0,m0}) = ( A*(i0,n0),
  A*(i0,m0) )][( A*(j0,n0),A*(j0,m0) )
proof
  assume that
A1: i0 < j0 and
A2: n0 < m0;
A3: card {n0,m0}=2 by A2,CARD_2:57;
A4: Sgm {n0,m0}=<*n0,m0*> by A2,FINSEQ_3:45;
  then
A5: Sgm {n0,m0}.1=n0;
A6: Sgm {i0,j0}=<*i0,j0*> by A1,FINSEQ_3:45;
  then
A7: Sgm {i0,j0}.1=i0;
A8: Sgm {i0,j0}.2=j0 by A6;
A9: Sgm {n0,m0}.2=m0 by A4;
  card {i0,j0}= 2 by A1,CARD_2:57;
  hence thesis by A3,A5,A9,A7,A8,Th23;
end;
