theorem Th34:
  D1 tolerates D2 & S IsNDRankSeq V,A & D1 in S.m & D2 in S.m
  implies D1 \/ D2 in S.m
  proof
    set D = D1\/D2;
    assume that
A1: D1 tolerates D2 and
A2: S IsNDRankSeq V,A and
A3: D1 in S.m & D2 in S.m;
A4: m in dom S by A3,FUNCT_1:def 2;
    not 0 in dom S by FINSEQ_3:24;
    then m <> 0 by A3,FUNCT_1:def 2;
    then
A5: 0+1 <= m by NAT_1:13;
    then reconsider z = m-1 as Element of NAT by INT_1:5;
    per cases by A5,XXREAL_0:1;
    suppose 1 < m;
      then
A6:   S.(z+1) = NDSS(V,A\/S.z) by A2,A4,CGAMES_1:20;
      then D1 is PartFunc of V,A\/S.z & D2 is PartFunc of V,A\/S.z by A3,Th4;
      then D is V-defined (A\/S.z)-valued;
      then D is PartFunc of V,A\/S.z by A1,RELSET_1:4,PARTFUN1:51;
      hence thesis by A6;
    end;
    suppose m = 1;
      hence thesis by A1,A2,A3,Th8;
    end;
  end;
