theorem Th23:
  t is x-omitting implies (t,[x,s])<-t1 = t
  proof assume
A1: Coim(t,[x,s]) = {};
    reconsider dt = dom t as set;
AA: dom ((t,[x,s])<-t1) = dt
    proof
      thus dom ((t,[x,s])<-t1) c= dt
      proof let a; assume a in dom ((t,[x,s])<-t1);
        then reconsider r = a as Node of (t,[x,s])<-t1;
        per cases by TREES_4:def 7;
        suppose r in dom t;
          hence thesis;
        end;
        suppose ex q being Node of t, p being Node of t1 st
          q in Leaves dom t & t.q = [x,s] & r = q^p;
          then consider q being Node of t, p being Node of t1 such that
A2:       q in Leaves dom t & t.q = [x,s] & r = q^p;
          [x,s] in {[x,s]} by TARSKI:def 1;
          hence thesis by A1,A2,FUNCT_1:def 7;
        end;
      end;
      thus dt c= dom ((t,[x,s])<-t1) by TREES_4:def 7;
    end;
    now let a; assume a in dom t;
      then reconsider r = a as Node of t;
      [x,s] in {[x,s]} by TARSKI:def 1;
      then t.r <> [x,s] by A1,FUNCT_1:def 7;
      hence ((t,[x,s])<-t1).a = t.a by TREES_4:def 7;
    end;
    hence thesis by AA,FUNCT_1:2;
  end;
