theorem Th70:
  p in q & q in s implies (T,p,q)incl.(p,s) = [q,s] & (T,p,q)incl.(q,s) = [p,s]
  proof assume
A1: p in q & q in s;
    set X = dom T;
    set i = id X;
    set f = Swap(i,p,q);
    set h = [:f,f:];
    set Y = (succ q)\p;
A2: dom i = X;
A3: s <> p & s <> q by A1;
    thus (T,p,q)incl.(p,s) = [f.p,f.s] by A1,Th65
    .= [f.p,i.s] by A3,Th33
    .= [f.p,s]
    .= [i.q,s] by A2,Th29
    .= [q,s];
    thus (T,p,q)incl.(q,s) = [f.q,f.s] by A1,Th65
    .= [f.q,i.s] by A3,Th33
    .= [f.q,s]
    .= [i.p,s] by A2,Th31
    .= [p,s];
  end;
