theorem Th45A:
  for s0 being SortSymbol of S, x0 being Element of X.s0 holds
  the_sort_of t = s & C is x0-omitting & t is x0-omitting implies
  C-sub t is x0-omitting
  proof
    let s0 be SortSymbol of S;
    let x0 be Element of X.s0;
    assume the_sort_of t = s;
    then
A1: C-sub t = (C,[x,s])<-t by SUB;
    assume Z1: Coim(C,[x0,s0]) = {};
    assume Z2: Coim(t,[x0,s0]) = {};
    assume Coim(C-sub t, [x0,s0]) <> {};
    then consider a such that
A2: a in Coim(C-sub t, [x0,s0]) by XBOOLE_0:7;
A3: a in dom(C-sub t) & (C-sub t).a in {[x0,s0]} by A2,FUNCT_1:def 7;
    reconsider a as Element of dom(C-sub t) by A2,FUNCT_1:def 7;
A5: now given q being Node of C, r being Node of t such that
B1:   q in Leaves dom C & C.q = [x,s] & a = q^r;
      (C-sub t).a = t.r by A1,B1,TREES_4:def 7;
      hence contradiction by Z2,A3,FUNCT_1:def 7;
    end;
    per cases by A1,TREES_4:def 7;
    suppose
B3:   a in dom C & C.a <> [x,s];
      then (C-sub t).a = C.a by A1,TREES_4:def 7;
      hence contradiction by Z1,A3,B3,FUNCT_1:def 7;
    end;
    suppose
B2:   a in dom C & C.a = [x,s];
      then reconsider q = a as Node of C;
      reconsider r = {} as Node of t by TREES_1:22;
      q^r = a & q in Leaves dom C by B2,Lem13;
      hence contradiction by A5,B2;
    end;
    suppose ex q being Node of C, r being Node of t st
      q in Leaves dom C & C.q = [x,s] & a = q^r;
      hence contradiction by A5;
    end;
  end;
