reserve A for RelStr;
reserve X for non empty set;
reserve PX,PY,PZ,Y,a,b,c,x,y for set;
reserve S1,S2 for Subset of Y;

theorem Th11:
  for B being mutually-disjoint Subset-Family of Y st for b be set
  st b in B holds S1 misses b & Y <> {} holds B \/ {S1} is mutually-disjoint
  Subset-Family of Y & (S1 <> {} implies union (B \/ {S1}) <> union B)
proof
  let B being mutually-disjoint Subset-Family of Y such that
A1: for b be set st b in B holds S1 misses b & Y <> {};
  set C = B \/ {S1};
A2: now
    let c1,c2 be set such that
A3: c1 in C and
A4: c2 in C and
A5: c1 <> c2;
    per cases by A3,XBOOLE_0:def 3;
    suppose
A6:   c1 in B;
      per cases by A4,XBOOLE_0:def 3;
      suppose
        c2 in B;
        hence c1 misses c2 by A5,A6,Def5;
      end;
      suppose
        c2 in {S1};
        then c2 = S1 by TARSKI:def 1;
        hence c1 misses c2 by A1,A6;
      end;
    end;
    suppose
      c1 in {S1};
      then
A7:   c1 = S1 by TARSKI:def 1;
      then not c2 in {S1} by A5,TARSKI:def 1;
      then c2 in B by A4,XBOOLE_0:def 3;
      hence c1 misses c2 by A1,A7;
    end;
  end;
  {S1} c= bool Y
  proof
    let p be object;
    assume p in {S1};
    then p = S1 by TARSKI:def 1;
    hence thesis;
  end;
  hence C is mutually-disjoint Subset-Family of Y by A2,Def5,XBOOLE_1:8;
  thus S1 <> {} implies union C <> union B
  proof
    assume
A8: S1 <> {};
    not union C c= union B
    proof
A9:   {S1} c= C by XBOOLE_1:7;
      assume
A10:  union C c= union B;
      consider p be object such that
A11:  p in S1 by A8,XBOOLE_0:def 1;
      S1 in {S1} by TARSKI:def 1;
      then p in union C by A11,A9,TARSKI:def 4;
      then consider S2 be set such that
A12:  p in S2 and
A13:  S2 in B by A10,TARSKI:def 4;
      S1 misses S2 by A1,A13;
      hence contradiction by A11,A12,XBOOLE_0:3;
    end;
    hence thesis;
  end;
end;
