theorem Th62:
  p c= r & r c= q implies
  (T,p,q)incl.(p,r) = [p, r] & (T,p,q)incl.(r,q) = [r,q]
  proof assume
A1: p c= r & r c= q;
    set Y = (succ q)\p;
    set Z1 = [:{p},Y:], Z2 = [:Y,{q}:];
    set g = id(Z1\/Z2);
    p in {p} & q in {q} & r in Y by A1,Th59,TARSKI:def 1; then
    [p,r] in Z1 & [r,q] in Z2 by ZFMISC_1:def 2; then
A2: [p,r] in Z1\/Z2 & [r,q] in Z1\/Z2 by XBOOLE_0:def 3;
A3: dom g = Z1\/Z2;
    hence (T,p,q)incl.(p,r) = g.(p,r) by A2,FUNCT_4:13
    .= [p, r] by A2,FUNCT_1:17;
    thus (T,p,q)incl.(r,q) = g.(r,q) by A2,A3,FUNCT_4:13
    .= [r, q] by A2,FUNCT_1:17;
  end;
