reserve U for Universe;

theorem
  for R being strict non empty RelStr st R in fin_RelStr_sp holds R is N-free
proof
  let R be strict non empty RelStr;
  set N4 = Necklace 4;
  defpred P[Nat] means ex S being strict non empty RelStr st S in
  fin_RelStr_sp & card (the carrier of S) = $1 & S embeds N4;
  assume
A1: R in fin_RelStr_sp;
  then
  ex S being strict RelStr st R = S & the carrier of S in FinSETS by Def4;
  then reconsider C = the carrier of R as Element of FinSETS;
  set k = card C;
A2: for m being Nat st m <> 0 & P[m] ex n being Nat st n < m & P[n]
  proof
    let m be Nat such that
    m <> 0 and
A3: P[m];
    consider S being non empty strict RelStr such that
A4: S in fin_RelStr_sp and
A5: card (the carrier of S) = m and
A6: S embeds N4 by A3;
    per cases by A4,Th6;
    suppose
A7:   S is strict 1-element RelStr;
A8:   dom the InternalRel of S c= (the carrier of S) by RELAT_1:def 18;
A9:   rng the InternalRel of S c= (the carrier of S) by RELAT_1:def 19;
      consider f being Function of N4,S such that
      f is one-to-one and
A10:  for x,y being Element of N4 holds [x,y] in the InternalRel of
      N4 iff [f.x,f.y] in the InternalRel of S by A6,NECKLACE:def 1;
A11:  the carrier of N4 = {0,1,2,3} by NECKLACE:1,20;
      then
A12:  0 in the carrier of N4 by ENUMSET1:def 2;
A13:  1 in the carrier of N4 by A11,ENUMSET1:def 2;
      0 = 0+1 or 1 = 0+1;
      then [0,1] in the InternalRel of N4 by A12,A13,NECKLACE:25;
      then
A14:  [f.0,f.1] in the InternalRel of S by A10,A12,A13;
      then
A15:  f.1 in rng (the InternalRel of S) by XTUPLE_0:def 13;
      f.0 in dom (the InternalRel of S) by A14,XTUPLE_0:def 12;
      then f.0 = f.1 by A7,A15,A8,A9,STRUCT_0:def 10;
      then [0,0] in (the InternalRel of N4) by A10,A12,A14;
      then
      [0,0] = [0,1] or [0,0]=[1,0] or [0,0]=[1,2] or [0,0]=[2,1] or [0,0]
      =[2,3] or [0,0]=[3,2] by Th2,ENUMSET1:def 4;
      hence thesis by XTUPLE_0:1;
    end;
    suppose
      ex H1,H2 being strict RelStr st the carrier of H1 misses the
carrier of H2 & H1 in fin_RelStr_sp & H2 in fin_RelStr_sp & (S = union_of(H1,H2
      ) or S = sum_of(H1,H2));
      then consider H1,H2 being strict RelStr such that
A16:  the carrier of H1 misses the carrier of H2 and
A17:  H1 in fin_RelStr_sp and
A18:  H2 in fin_RelStr_sp and
A19:  S = union_of(H1,H2) or S = sum_of(H1,H2);
A20:  now
A21:    not [1,3] in the InternalRel of N4
        proof
          assume
A22:      [1,3] in the InternalRel of N4;
          per cases by A22,Th2,ENUMSET1:def 4;
          suppose
            [1,3] = [0,1];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [1,3] = [1,2];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [1,3] = [2,3];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [1,3] = [3,2];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [1,3] = [2,1];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [1,3] = [1,0];
            hence contradiction by XTUPLE_0:1;
          end;
        end;
A23:    not [0,2] in the InternalRel of N4
        proof
          assume
A24:      [0,2] in the InternalRel of N4;
          per cases by A24,Th2,ENUMSET1:def 4;
          suppose
            [0,2] = [0,1];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [0,2] = [1,2];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [0,2] = [2,3];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [0,2] = [3,2];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [0,2] = [2,1];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [0,2] = [1,0];
            hence contradiction by XTUPLE_0:1;
          end;
        end;
A25:    the carrier of N4 = {0,1,2,3} by NECKLACE:1,20;
        then
A26:    0 in the carrier of N4 by ENUMSET1:def 2;
A27:    not [0,3] in the InternalRel of N4
        proof
          assume
A28:      [0,3] in the InternalRel of N4;
          per cases by A28,Th2,ENUMSET1:def 4;
          suppose
            [0,3] = [0,1];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [0,3] = [1,2];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [0,3] = [2,3];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [0,3] = [3,2];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [0,3] = [2,1];
            hence contradiction by XTUPLE_0:1;
          end;
          suppose
            [0,3] = [1,0];
            hence contradiction by XTUPLE_0:1;
          end;
        end;
        set A = the InternalRel of H1, B = the InternalRel of H2, C = [:the
carrier of H1,the carrier of H2:], D = [:the carrier of H2, the carrier of H1:]
, cH1 = the carrier of H1, cH2 = the carrier of H2, cS = the carrier of S;
A29:    dom (the InternalRel of S) c= cS by RELAT_1:def 18;
        assume
A30:    S = sum_of(H1,H2);
A31:    C c= (the InternalRel of S)
        proof
          let x be object;
          assume x in C;
          then
A32:      x in (A \/ B) \/ C by XBOOLE_0:def 3;
          ((A \/ B) \/ C) c= ((A \/ B) \/ C) \/ D by XBOOLE_1:7;
          then ((A \/ B) \/ C) c= the InternalRel of S by A30,Def3;
          hence thesis by A32;
        end;
A33:    rng (the InternalRel of S) c= cS by RELAT_1:def 19;
A34:    3 in the carrier of N4 by A25,ENUMSET1:def 2;
A35:    D c= (the InternalRel of S)
        proof
          let x be object;
          ((B \/ C) \/ D) c= A \/ ((B \/ C) \/ D) by XBOOLE_1:7;
          then ((B \/ C) \/ D) c= A \/ (B \/ (C \/ D)) by XBOOLE_1:4;
          then ((B \/ C) \/ D) c= (A \/ B) \/ (C \/ D) by XBOOLE_1:4;
          then ((B \/ C) \/ D) c= ((A \/ B) \/ C) \/ D by XBOOLE_1:4;
          then
A36:      ((B \/ C) \/ D) c= the InternalRel of S by A30,Def3;
          assume x in D;
          then x in (B \/ C) \/ D by XBOOLE_0:def 3;
          hence thesis by A36;
        end;
A37:    rng A c= cH1 by RELAT_1:def 19;
A38:    1 in the carrier of N4 by A25,ENUMSET1:def 2;
        consider f being Function of N4,S such that
A39:    f is one-to-one and
A40:    for x,y being Element of N4 holds [x,y] in the InternalRel
        of N4 iff [f.x,f.y] in the InternalRel of S by A6,NECKLACE:def 1;
        [1,0] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A41:    [f.1,f.0] in the InternalRel of S by A40,A26,A38;
A42:    2 in the carrier of N4 by A25,ENUMSET1:def 2;
        [3,2] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A43:    [f.3,f.2] in the InternalRel of S by A40,A42,A34;
        [2,1] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A44:    [f.2,f.1] in the InternalRel of S by A40,A38,A42;
A45:    rng B c= cH2 by RELAT_1:def 19;
        [2,3] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A46:    [f.2,f.3] in the InternalRel of S by A40,A42,A34;
        then f.3 in rng (the InternalRel of S) by XTUPLE_0:def 13;
        then f.3 in cS by A33;
        then
A47:    f.3 in cH1 \/ cH2 by A30,Def3;
        [0,1] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A48:    [f.0,f.1] in the InternalRel of S by A40,A26,A38;
        then f.0 in dom (the InternalRel of S) by XTUPLE_0:def 12;
        then f.0 in cS by A29;
        then
A49:    f.0 in cH1 \/ cH2 by A30,Def3;
        f.1 in rng (the InternalRel of S) by A48,XTUPLE_0:def 13;
        then f.1 in cS by A33;
        then
A50:    f.1 in cH1 \/ cH2 by A30,Def3;
A51:    dom A c= cH1 by RELAT_1:def 18;
        [1,2] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A52:    [f.1,f.2] in the InternalRel of S by A40,A38,A42;
        then f.2 in rng (the InternalRel of S) by XTUPLE_0:def 13;
        then f.2 in cS by A33;
        then
A53:    f.2 in cH1 \/ cH2 by A30,Def3;
A54:    dom B c= cH2 by RELAT_1:def 18;
        per cases by A49,XBOOLE_0:def 3;
        suppose
A55:      f.0 in cH1;
          set x1=[f.0,f.1], x2=[f.1,f.2], x3=[f.2,f.3], x4=[f.1,f.0], x5=[f.2,
          f.1], x6=[f.3,f.2];
A56:      now
            assume x1 in D;
            then f.0 in cH2 by ZFMISC_1:87;
            hence contradiction by A16,A55,XBOOLE_0:3;
          end;
A57:      now
            assume f.2 in cH2;
            then [f.0,f.2] in C by A55,ZFMISC_1:87;
            hence contradiction by A40,A26,A42,A23,A31;
          end;
A58:      now
            assume x4 in B;
            then f.0 in rng B by XTUPLE_0:def 13;
            hence contradiction by A16,A45,A55,XBOOLE_0:3;
          end;
A59:      now
            assume x4 in C;
            then f.0 in cH2 by ZFMISC_1:87;
            hence contradiction by A16,A55,XBOOLE_0:3;
          end;
A60:      now
            assume x1 in B;
            then f.0 in dom B by XTUPLE_0:def 12;
            hence contradiction by A16,A54,A55,XBOOLE_0:3;
          end;
A61:      now
            assume f.3 in cH2;
            then [f.0,f.3] in C by A55,ZFMISC_1:87;
            hence contradiction by A40,A26,A34,A27,A31;
          end;
          then
A62:      f.3 in cH1 by A47,XBOOLE_0:def 3;
A63:      now
            assume f.1 in cH2;
            then [f.1,f.3] in D by A62,ZFMISC_1:87;
            hence contradiction by A40,A38,A34,A21,A35;
          end;
A64:      dom f = the carrier of N4 by FUNCT_2:def 1;
          rng f c= the carrier of H1
          proof
            let y be object;
            assume y in rng f;
            then consider x being object such that
A65:        x in dom f and
A66:        y = f.x by FUNCT_1:def 3;
            per cases by A64,A65,Lm1,ENUMSET1:def 2;
            suppose
              x = 0;
              hence thesis by A55,A66;
            end;
            suppose
              x = 1;
              hence thesis by A50,A63,A66,XBOOLE_0:def 3;
            end;
            suppose
              x = 2;
              hence thesis by A53,A57,A66,XBOOLE_0:def 3;
            end;
            suppose
              x = 3;
              hence thesis by A47,A61,A66,XBOOLE_0:def 3;
            end;
          end;
          then
A67:      f is Function of the carrier of N4,cH1 by FUNCT_2:6;
          H1 is finite by A17,Th4;
          then reconsider cH1 as finite set;
A68:      H1 is strict non empty RelStr by A17,Th4;
A69:      now
            assume x6 in B;
            then f.2 in rng B by XTUPLE_0:def 13;
            hence contradiction by A45,A57;
          end;
A70:      now
            assume x3 in B;
            then f.2 in dom B by XTUPLE_0:def 12;
            hence contradiction by A54,A57;
          end;
          x4 in (A \/ B \/ C \/ D) by A30,A41,Def3;
          then x4 in ((A\/ B) \/ C) or x4 in D by XBOOLE_0:def 3;
          then x4 in (A \/ B) or x4 in C or x4 in D by XBOOLE_0:def 3;
          then
A71:      [f.1,f.0] in A by A58,A59,A63,XBOOLE_0:def 3,ZFMISC_1:87;
A72:      now
            assume x2 in B;
            then f.1 in dom B by XTUPLE_0:def 12;
            hence contradiction by A54,A63;
          end;
          x2 in (A \/ B \/ C \/ D) by A30,A52,Def3;
          then x2 in ((A\/ B) \/ C) or x2 in D by XBOOLE_0:def 3;
          then x2 in (A \/ B) or x2 in C or x2 in D by XBOOLE_0:def 3;
          then
A73:      [f.1,f.2] in A by A72,A57,A63,XBOOLE_0:def 3,ZFMISC_1:87;
          x6 in (A \/ B \/ C \/ D) by A30,A43,Def3;
          then x6 in ((A\/ B) \/ C) or x6 in D by XBOOLE_0:def 3;
          then x6 in (A \/ B) or x6 in C or x6 in D by XBOOLE_0:def 3;
          then
A74:      [f.3,f.2] in A by A69,A57,A61,XBOOLE_0:def 3,ZFMISC_1:87;
          x3 in (A \/ B \/ C \/ D) by A30,A46,Def3;
          then x3 in ((A\/ B) \/ C) or x3 in D by XBOOLE_0:def 3;
          then x3 in (A \/ B) or x3 in C or x3 in D by XBOOLE_0:def 3;
          then
A75:      [f.2,f.3] in A by A70,A61,A57,XBOOLE_0:def 3,ZFMISC_1:87;
A76:      now
            assume x5 in B;
            then f.1 in rng B by XTUPLE_0:def 13;
            hence contradiction by A45,A63;
          end;
          H2 is finite by A18,Th4;
          then reconsider cH2 as finite set;
          cS = cH1 \/ cH2 by A30,Def3;
          then reconsider cS as finite set;
A77:      H2 is non empty by A18,Th4;
A78:      cH1 <> cS
          proof
A79:        cS = cH1 \/ cH2 by A30,Def3;
            assume
A80:        cH1 = cS;
            consider x being object such that
A81:        x in cH2 by A77,XBOOLE_0:def 1;
            cH1 /\ cH2 = {} by A16,XBOOLE_0:def 7;
            then not x in cH1 by A81,XBOOLE_0:def 4;
            hence contradiction by A80,A79,A81,XBOOLE_0:def 3;
          end;
          cS = cH1 \/ cH2 by A30,Def3;
          then cH1 c= cS by XBOOLE_1:7;
          then cH1 c< cS by A78,XBOOLE_0:def 8;
          then
A82:      card cH1 < card cS by CARD_2:48;
          x5 in (A \/ B \/ C \/ D) by A30,A44,Def3;
          then x5 in ((A\/ B) \/ C) or x5 in D by XBOOLE_0:def 3;
          then x5 in (A \/ B) or x5 in C or x5 in D by XBOOLE_0:def 3;
          then
A83:      [f.2,f.1] in A by A76,A63,A57,XBOOLE_0:def 3,ZFMISC_1:87;
          x1 in (A \/ B \/ C \/ D) by A30,A48,Def3;
          then x1 in ((A \/ B) \/ C ) or x1 in D by XBOOLE_0:def 3;
          then x1 in (A \/ B) or x1 in C or x1 in D by XBOOLE_0:def 3;
          then
A84:      [f.0,f.1] in A by A60,A63,A56,XBOOLE_0:def 3,ZFMISC_1:87;
          for x,y being Element of N4 holds [x,y] in the InternalRel of
          N4 iff [f.x,f.y] in the InternalRel of H1
          proof
            let x,y being Element of N4;
            thus [x,y] in the InternalRel of N4 implies [f.x,f.y] in the
            InternalRel of H1
            proof
              assume
A85:          [x,y] in the InternalRel of N4;
              per cases by A85,Th2,ENUMSET1:def 4;
              suppose
A86:            [x,y] = [0,1];
                then x = 0 by XTUPLE_0:1;
                hence thesis by A84,A86,XTUPLE_0:1;
              end;
              suppose
A87:            [x,y] = [1,0];
                then x = 1 by XTUPLE_0:1;
                hence thesis by A71,A87,XTUPLE_0:1;
              end;
              suppose
A88:            [x,y] = [1,2];
                then x = 1 by XTUPLE_0:1;
                hence thesis by A73,A88,XTUPLE_0:1;
              end;
              suppose
A89:            [x,y] = [2,1];
                then x = 2 by XTUPLE_0:1;
                hence thesis by A83,A89,XTUPLE_0:1;
              end;
              suppose
A90:           [x,y] = [2,3];
                then x = 2 by XTUPLE_0:1;
                hence thesis by A75,A90,XTUPLE_0:1;
              end;
              suppose
A91:           [x,y] = [3,2];
                then x = 3 by XTUPLE_0:1;
                hence thesis by A74,A91,XTUPLE_0:1;
              end;
            end;
            thus [f.x,f.y] in the InternalRel of H1 implies [x,y] in the
            InternalRel of N4
            proof
              [f.x,f.y] in the InternalRel of S implies [x,y] in the
              InternalRel of N4 by A40;
              then [f.x,f.y] in (((A \/ B) \/ C) \/ D) implies [x,y] in the
              InternalRel of N4 by A30,Def3;
              then [f.x,f.y] in ((A \/ B) \/ (C\/ D)) implies [x,y] in the
              InternalRel of N4 by XBOOLE_1:4;
              then
A92:         [f.x,f.y] in (A \/ (B\/(C\/D))) implies [x,y] in the
              InternalRel of N4 by XBOOLE_1:4;
              assume [f.x,f.y] in the InternalRel of H1;
              hence thesis by A92,XBOOLE_0:def 3;
            end;
          end;
          then H1 embeds N4 by A39,A67,NECKLACE:def 1;
          hence thesis by A5,A17,A68,A82;
        end;
        suppose
A93:     f.0 in the carrier of H2;
          set x1=[f.0,f.1], x2=[f.1,f.2], x3=[f.2,f.3], x4=[f.1,f.0], x5=[f.2,
          f.1], x6=[f.3,f.2];
A94:     now
            assume x1 in C;
            then f.0 in cH1 by ZFMISC_1:87;
            hence contradiction by A16,A93,XBOOLE_0:3;
          end;
A95:     now
            assume x4 in D;
            then f.0 in cH1 by ZFMISC_1:87;
            hence contradiction by A16,A93,XBOOLE_0:3;
          end;
A96:     now
            assume f.3 in cH1;
            then [f.0,f.3] in D by A93,ZFMISC_1:87;
            hence contradiction by A40,A26,A34,A27,A35;
          end;
          then
A97:     f.3 in cH2 by A47,XBOOLE_0:def 3;
A98:     now
            assume f.1 in cH1;
            then [f.1,f.3] in C by A97,ZFMISC_1:87;
            hence contradiction by A40,A38,A34,A21,A31;
          end;
A99:     now
            assume x4 in A;
            then f.1 in dom A by XTUPLE_0:def 12;
            hence contradiction by A51,A98;
          end;
          x4 in (A \/ B \/ C \/ D) by A30,A41,Def3;
          then x4 in ((A\/ B) \/ C) or x4 in D by XBOOLE_0:def 3;
          then x4 in (A \/ B) or x4 in C or x4 in D by XBOOLE_0:def 3;
          then
A100:     [f.1,f.0] in B by A99,A98,A95,XBOOLE_0:def 3,ZFMISC_1:87;
A101:     now
            assume x5 in A;
            then f.1 in rng A by XTUPLE_0:def 13;
            hence contradiction by A37,A98;
          end;
A102:     now
            assume f.2 in cH1;
            then [f.0,f.2] in D by A93,ZFMISC_1:87;
            hence contradiction by A40,A26,A42,A23,A35;
          end;
A103:     dom f = the carrier of N4 by FUNCT_2:def 1;
          rng f c= the carrier of H2
          proof
            let y be object;
            assume y in rng f;
            then consider x being object such that
A104:       x in dom f and
A105:       y = f.x by FUNCT_1:def 3;
            per cases by A103,A104,Lm1,ENUMSET1:def 2;
            suppose
              x = 0;
              hence thesis by A93,A105;
            end;
            suppose
              x = 1;
              hence thesis by A50,A98,A105,XBOOLE_0:def 3;
            end;
            suppose
              x = 2;
              hence thesis by A53,A102,A105,XBOOLE_0:def 3;
            end;
            suppose
              x = 3;
              hence thesis by A47,A96,A105,XBOOLE_0:def 3;
            end;
          end;
          then
A106:     f is Function of the carrier of N4,cH2 by FUNCT_2:6;
          H1 is finite by A17,Th4;
          then reconsider cH1 as finite set;
A107:     H2 is strict non empty RelStr by A18,Th4;
A108:     now
            assume x1 in A;
            then f.0 in dom A by XTUPLE_0:def 12;
            hence contradiction by A16,A51,A93,XBOOLE_0:3;
          end;
A109:     now
            assume x6 in A;
            then f.2 in rng A by XTUPLE_0:def 13;
            hence contradiction by A37,A102;
          end;
A110:     now
            assume x3 in A;
            then f.2 in dom A by XTUPLE_0:def 12;
            hence contradiction by A51,A102;
          end;
          x6 in (A \/ B \/ C \/ D) by A30,A43,Def3;
          then x6 in ((A\/ B) \/ C) or x6 in D by XBOOLE_0:def 3;
          then x6 in (A \/ B) or x6 in C or x6 in D by XBOOLE_0:def 3;
          then
A111:     [f.3,f.2] in B by A109,A96,A102,XBOOLE_0:def 3,ZFMISC_1:87;
A112:     now
            assume x2 in A;
            then f.2 in rng A by XTUPLE_0:def 13;
            hence contradiction by A37,A102;
          end;
          x2 in (A \/ B \/ C \/ D) by A30,A52,Def3;
          then x2 in ((A\/ B) \/ C) or x2 in D by XBOOLE_0:def 3;
          then x2 in (A \/ B) or x2 in C or x2 in D by XBOOLE_0:def 3;
          then
A113:     [f.1,f.2] in B by A112,A98,A102,XBOOLE_0:def 3,ZFMISC_1:87;
          x3 in (A \/ B \/ C \/ D) by A30,A46,Def3;
          then x3 in ((A\/ B) \/ C) or x3 in D by XBOOLE_0:def 3;
          then x3 in (A \/ B) or x3 in C or x3 in D by XBOOLE_0:def 3;
          then
A114:     [f.2,f.3] in B by A110,A102,A96,XBOOLE_0:def 3,ZFMISC_1:87;
          x5 in (A \/ B \/ C \/ D) by A30,A44,Def3;
          then x5 in ((A\/ B) \/ C) or x5 in D by XBOOLE_0:def 3;
          then x5 in (A \/ B) or x5 in C or x5 in D by XBOOLE_0:def 3;
          then
A115:     [f.2,f.1] in B by A101,A102,A98,XBOOLE_0:def 3,ZFMISC_1:87;
          H2 is finite by A18,Th4;
          then reconsider cH2 as finite set;
          cS = cH1 \/ cH2 by A30,Def3;
          then reconsider cS as finite set;
A116:     H1 is non empty by A17,Th4;
A117:     cH2 <> cS
          proof
A118:       cS = cH1 \/ cH2 by A30,Def3;
            assume
A119:       cH2 = cS;
            consider x being object such that
A120:       x in cH1 by A116,XBOOLE_0:def 1;
            cH1 /\ cH2 = {} by A16,XBOOLE_0:def 7;
            then not x in cH2 by A120,XBOOLE_0:def 4;
            hence contradiction by A119,A118,A120,XBOOLE_0:def 3;
          end;
          cS = cH1 \/ cH2 by A30,Def3;
          then cH2 c= cS by XBOOLE_1:7;
          then cH2 c< cS by A117,XBOOLE_0:def 8;
          then
A121:     card cH2 < card cS by CARD_2:48;
          x1 in (A \/ B \/ C \/ D) by A30,A48,Def3;
          then x1 in ((A \/ B) \/ C ) or x1 in D by XBOOLE_0:def 3;
          then x1 in (A \/ B) or x1 in C or x1 in D by XBOOLE_0:def 3;
          then
A122:     [f.0,f.1] in B by A108,A94,A98,XBOOLE_0:def 3,ZFMISC_1:87;
          for x,y being Element of N4 holds [x,y] in the InternalRel of
          N4 iff [f.x,f.y] in the InternalRel of H2
          proof
            let x,y being Element of N4;
            thus [x,y] in the InternalRel of N4 implies [f.x,f.y] in the
            InternalRel of H2
            proof
              assume
A123:         [x,y] in the InternalRel of N4;
              per cases by A123,Th2,ENUMSET1:def 4;
              suppose
A124:           [x,y] = [0,1];
                then x = 0 by XTUPLE_0:1;
                hence thesis by A122,A124,XTUPLE_0:1;
              end;
              suppose
A125:           [x,y] = [1,0];
                then x = 1 by XTUPLE_0:1;
                hence thesis by A100,A125,XTUPLE_0:1;
              end;
              suppose
A126:           [x,y] = [1,2];
                then x = 1 by XTUPLE_0:1;
                hence thesis by A113,A126,XTUPLE_0:1;
              end;
              suppose
A127:           [x,y] = [2,1];
                then x = 2 by XTUPLE_0:1;
                hence thesis by A115,A127,XTUPLE_0:1;
              end;
              suppose
A128:           [x,y] = [2,3];
                then x = 2 by XTUPLE_0:1;
                hence thesis by A114,A128,XTUPLE_0:1;
              end;
              suppose
A129:           [x,y] = [3,2];
                then x = 3 by XTUPLE_0:1;
                hence thesis by A111,A129,XTUPLE_0:1;
              end;
            end;
            thus [f.x,f.y] in the InternalRel of H2 implies [x,y] in the
            InternalRel of N4
            proof
              [f.x,f.y] in the InternalRel of S implies [x,y] in the
              InternalRel of N4 by A40;
              then [f.x,f.y] in (((A \/ B) \/ C) \/ D) implies [x,y] in the
              InternalRel of N4 by A30,Def3;
              then [f.x,f.y] in ((A \/ B) \/ (C\/ D)) implies [x,y] in the
              InternalRel of N4 by XBOOLE_1:4;
              then
A130:         [f.x,f.y] in (B \/ (A\/(C\/D))) implies [x,y] in the
              InternalRel of N4 by XBOOLE_1:4;
              assume [f.x,f.y] in the InternalRel of H2;
              hence thesis by A130,XBOOLE_0:def 3;
            end;
          end;
          then H2 embeds N4 by A39,A106,NECKLACE:def 1;
          hence thesis by A5,A18,A107,A121;
        end;
      end;
      now
A131:   the carrier of N4 = {0,1,2,3} by NECKLACE:1,20;
        then
A132:   0 in the carrier of N4 by ENUMSET1:def 2;
A133:   3 in the carrier of N4 by A131,ENUMSET1:def 2;
A134:   1 in the carrier of N4 by A131,ENUMSET1:def 2;
A135:   dom (the InternalRel of S) c= (the carrier of S) by RELAT_1:def 18;
        consider f being Function of N4,S such that
A136:   f is one-to-one and
A137:   for x,y being Element of N4 holds [x,y] in the InternalRel of
        N4 iff [f.x,f.y] in the InternalRel of S by A6,NECKLACE:def 1;
        assume
A138:   S = union_of(H1,H2);
        [0,1] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A139:   [f.0,f.1] in the InternalRel of S by A137,A132,A134;
        then f.0 in dom (the InternalRel of S) by XTUPLE_0:def 12;
        then f.0 in (the carrier of S) by A135;
        then
A140:   f.0 in (the carrier of H1) \/ (the carrier of H2) by A138,Def2;
A141:   2 in the carrier of N4 by A131,ENUMSET1:def 2;
        [3,2] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A142:   [f.3,f.2] in the InternalRel of S by A137,A141,A133;
        [2,3] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A143:   [f.2,f.3] in the InternalRel of S by A137,A141,A133;
        [2,1] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A144:   [f.2,f.1] in the InternalRel of S by A137,A134,A141;
        [1,2] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A145:   [f.1,f.2] in the InternalRel of S by A137,A134,A141;
        [1,0] in the InternalRel of N4 by Th2,ENUMSET1:def 4;
        then
A146:   [f.1,f.0] in the InternalRel of S by A137,A132,A134;
        per cases by A140,XBOOLE_0:def 3;
        suppose
A147:     f.0 in the carrier of H1;
          set cS = the carrier of S, cH1 = the carrier of H1, cH2 = the
          carrier of H2;
A148:     dom f = the carrier of N4 by FUNCT_2:def 1;
          H2 is finite by A18,Th4;
          then reconsider cH2 as finite set;
          H1 is finite by A17,Th4;
          then reconsider cH1 as finite set;
A149:     cS = cH1 \/ cH2 by A138,Def2;
A150:     dom (the InternalRel of H2) c= (the carrier of H2) by RELAT_1:def 18;
A151:     now
            assume [f.0,f.1] in the InternalRel of H2;
            then f.0 in dom (the InternalRel of H2) by XTUPLE_0:def 12;
            hence contradiction by A16,A147,A150,XBOOLE_0:3;
          end;
A152:     rng (the InternalRel of H2) c= (the carrier of H2) by RELAT_1:def 19;
A153:     now
            assume [f.1,f.0] in the InternalRel of H2;
            then f.0 in rng (the InternalRel of H2) by XTUPLE_0:def 13;
            hence contradiction by A16,A147,A152,XBOOLE_0:3;
          end;
          [f.1,f.0] in (the InternalRel of H1) \/ (the InternalRel of H2)
          by A138,A146,Def2;
          then
A154:     [f.1,f.0] in the InternalRel of H1 by A153,XBOOLE_0:def 3;
A155:     H1 is strict non empty RelStr by A17,Th4;
          reconsider cS as finite set by A149;
A156:     rng (the InternalRel of H1) c= (the carrier of H1) by RELAT_1:def 19;
          [f.0,f.1] in (the InternalRel of H1) \/ (the InternalRel of H2)
          by A138,A139,Def2;
          then
A157:     [f.0,f.1] in the InternalRel of H1 by A151,XBOOLE_0:def 3;
          then
A158:     f.1 in rng (the InternalRel of H1) by XTUPLE_0:def 13;
A159:     now
            assume [f.2,f.1] in the InternalRel of H2;
            then f.1 in rng (the InternalRel of H2) by XTUPLE_0:def 13;
            hence contradiction by A16,A156,A152,A158,XBOOLE_0:3;
          end;
A160:     H2 is non empty by A18,Th4;
A161:     cH1 <> cS
          proof
A162:       cS = cH1 \/ cH2 by A138,Def2;
            assume
A163:       cH1 = cS;
            consider x being object such that
A164:       x in cH2 by A160,XBOOLE_0:def 1;
            cH1 /\ cH2 = {} by A16,XBOOLE_0:def 7;
            then not x in cH1 by A164,XBOOLE_0:def 4;
            hence contradiction by A163,A162,A164,XBOOLE_0:def 3;
          end;
          cS = cH1 \/ cH2 by A138,Def2;
          then cH1 c= cS by XBOOLE_1:7;
          then cH1 c< cS by A161,XBOOLE_0:def 8;
          then
A165:     card cH1 < card cS by CARD_2:48;
A166:     [f.2,f.3] in (the InternalRel of H1) \/ (the InternalRel of H2)
          by A138,A143,Def2;
A167:     [f.1,f.2] in (the InternalRel of H1) \/ (the InternalRel of H2)
          by A138,A145,Def2;
          now
            assume [f.1,f.2] in the InternalRel of H2;
            then f.1 in dom (the InternalRel of H2) by XTUPLE_0:def 12;
            hence contradiction by A16,A156,A150,A158,XBOOLE_0:3;
          end;
          then
A168:     [f.1,f.2] in the InternalRel of H1 by A167,XBOOLE_0:def 3;
          then
A169:     f.2 in rng (the InternalRel of H1) by XTUPLE_0:def 13;
          now
            assume [f.2,f.3] in the InternalRel of H2;
            then f.2 in dom (the InternalRel of H2) by XTUPLE_0:def 12;
            hence contradiction by A16,A156,A150,A169,XBOOLE_0:3;
          end;
          then
A170:     [f.2,f.3] in the InternalRel of H1 by A166,XBOOLE_0:def 3;
          [f.2,f.1] in (the InternalRel of H1) \/ (the InternalRel of H2)
          by A138,A144,Def2;
          then
A171:     [f.2,f.1] in the InternalRel of H1 by A159,XBOOLE_0:def 3;
A172:     now
            assume [f.3,f.2] in the InternalRel of H2;
            then f.2 in rng (the InternalRel of H2) by XTUPLE_0:def 13;
            hence contradiction by A16,A156,A152,A169,XBOOLE_0:3;
          end;
          [f.3,f.2] in (the InternalRel of H1) \/ (the InternalRel of H2)
          by A138,A142,Def2;
          then
A173:     [f.3,f.2] in the InternalRel of H1 by A172,XBOOLE_0:def 3;
A174:     for x,y being Element of N4 holds [x,y] in the InternalRel of
          N4 iff [f.x,f.y] in the InternalRel of H1
          proof
            let x,y being Element of N4;
            thus [x,y] in the InternalRel of N4 implies [f.x,f.y] in the
            InternalRel of H1
            proof
              assume
A175:         [x,y] in the InternalRel of N4;
              per cases by A175,Th2,ENUMSET1:def 4;
              suppose
A176:           [x,y] = [0,1];
                then x = 0 by XTUPLE_0:1;
                hence thesis by A157,A176,XTUPLE_0:1;
              end;
              suppose
A177:           [x,y] = [1,0];
                then x = 1 by XTUPLE_0:1;
                hence thesis by A154,A177,XTUPLE_0:1;
              end;
              suppose
A178:           [x,y] = [1,2];
                then x = 1 by XTUPLE_0:1;
                hence thesis by A168,A178,XTUPLE_0:1;
              end;
              suppose
A179:           [x,y] = [2,1];
                then x = 2 by XTUPLE_0:1;
                hence thesis by A171,A179,XTUPLE_0:1;
              end;
              suppose
A180:           [x,y] = [2,3];
                then x = 2 by XTUPLE_0:1;
                hence thesis by A170,A180,XTUPLE_0:1;
              end;
              suppose
A181:           [x,y] = [3,2];
                then x = 3 by XTUPLE_0:1;
                hence thesis by A173,A181,XTUPLE_0:1;
              end;
            end;
            thus [f.x,f.y] in the InternalRel of H1 implies [x,y] in the
            InternalRel of N4
            proof
              [f.x,f.y] in the InternalRel of S implies [x,y] in the
              InternalRel of N4 by A137;
              then
A182:         [f.x,f.y] in (the InternalRel of H1) \/ (the InternalRel
              of H2) implies [x,y] in the InternalRel of N4 by A138,Def2;
              assume [f.x,f.y] in the InternalRel of H1;
              hence thesis by A182,XBOOLE_0:def 3;
            end;
          end;
A183:     f.3 in rng (the InternalRel of H1) by A170,XTUPLE_0:def 13;
          rng f c= the carrier of H1
          proof
            let y be object;
            assume y in rng f;
            then consider x being object such that
A184:       x in dom f and
A185:       y = f.x by FUNCT_1:def 3;
            per cases by A148,A184,Lm1,ENUMSET1:def 2;
            suppose
              x = 0;
              hence thesis by A147,A185;
            end;
            suppose
              x = 1;
              hence thesis by A156,A158,A185;
            end;
            suppose
              x = 2;
              hence thesis by A156,A169,A185;
            end;
            suppose
              x = 3;
              hence thesis by A156,A183,A185;
            end;
          end;
          then f is Function of the carrier of N4,the carrier of H1 by
FUNCT_2:6;
          then H1 embeds N4 by A136,A174,NECKLACE:def 1;
          hence thesis by A5,A17,A155,A165;
        end;
        suppose
A186:     f.0 in the carrier of H2;
          set cS = the carrier of S, cH1 = the carrier of H1, cH2 = the
          carrier of H2;
A187:     dom f = the carrier of N4 by FUNCT_2:def 1;
          H2 is finite by A18,Th4;
          then reconsider cH2 as finite set;
          H1 is finite by A17,Th4;
          then reconsider cH1 as finite set;
A188:     cS = cH1 \/ cH2 by A138,Def2;
A189:     dom (the InternalRel of H1) c= (the carrier of H1) by RELAT_1:def 18;
A190:     now
            assume [f.0,f.1] in the InternalRel of H1;
            then f.0 in dom (the InternalRel of H1) by XTUPLE_0:def 12;
            hence contradiction by A16,A186,A189,XBOOLE_0:3;
          end;
A191:     rng (the InternalRel of H1) c= (the carrier of H1) by RELAT_1:def 19;
A192:     now
            assume [f.1,f.0] in the InternalRel of H1;
            then f.0 in rng (the InternalRel of H1) by XTUPLE_0:def 13;
            hence contradiction by A16,A186,A191,XBOOLE_0:3;
          end;
          [f.1,f.0] in (the InternalRel of H1) \/ (the InternalRel of H2
          ) by A138,A146,Def2;
          then
A193:     [f.1,f.0] in the InternalRel of H2 by A192,XBOOLE_0:def 3;
A194:     H2 is strict non empty RelStr by A18,Th4;
          reconsider cS as finite set by A188;
A195:     rng (the InternalRel of H2) c= (the carrier of H2) by RELAT_1:def 19;
          [f.0,f.1] in (the InternalRel of H1) \/ (the InternalRel of H2
          ) by A138,A139,Def2;
          then
A196:     [f.0,f.1] in the InternalRel of H2 by A190,XBOOLE_0:def 3;
          then
A197:     f.1 in rng (the InternalRel of H2) by XTUPLE_0:def 13;
A198:     now
            assume [f.2,f.1] in the InternalRel of H1;
            then f.1 in rng (the InternalRel of H1) by XTUPLE_0:def 13;
            hence contradiction by A16,A195,A191,A197,XBOOLE_0:3;
          end;
A199:     H1 is non empty by A17,Th4;
A200:     cH2 <> cS
          proof
A201:       cS = cH1 \/ cH2 by A138,Def2;
            assume
A202:       cH2 = cS;
            consider x being object such that
A203:       x in cH1 by A199,XBOOLE_0:def 1;
            cH1 /\ cH2 = {} by A16,XBOOLE_0:def 7;
            then not x in cH2 by A203,XBOOLE_0:def 4;
            hence contradiction by A202,A201,A203,XBOOLE_0:def 3;
          end;
          cS = cH1 \/ cH2 by A138,Def2;
          then cH2 c= cS by XBOOLE_1:7;
          then cH2 c< cS by A200,XBOOLE_0:def 8;
          then
A204:     card cH2 < card cS by CARD_2:48;
A205:     [f.2,f.3] in (the InternalRel of H1) \/ (the InternalRel of H2
          ) by A138,A143,Def2;
A206:     [f.1,f.2] in (the InternalRel of H1) \/ (the InternalRel of H2
          ) by A138,A145,Def2;
          now
            assume [f.1,f.2] in the InternalRel of H1;
            then f.1 in dom (the InternalRel of H1) by XTUPLE_0:def 12;
            hence contradiction by A16,A195,A189,A197,XBOOLE_0:3;
          end;
          then
A207:     [f.1,f.2] in the InternalRel of H2 by A206,XBOOLE_0:def 3;
          then
A208:     f.2 in rng (the InternalRel of H2) by XTUPLE_0:def 13;
          now
            assume [f.2,f.3] in the InternalRel of H1;
            then f.2 in dom (the InternalRel of H1) by XTUPLE_0:def 12;
            hence contradiction by A16,A195,A189,A208,XBOOLE_0:3;
          end;
          then
A209:     [f.2,f.3] in the InternalRel of H2 by A205,XBOOLE_0:def 3;
          [f.2,f.1] in (the InternalRel of H1) \/ (the InternalRel of H2
          ) by A138,A144,Def2;
          then
A210:     [f.2,f.1] in the InternalRel of H2 by A198,XBOOLE_0:def 3;
A211:     now
            assume [f.3,f.2] in the InternalRel of H1;
            then f.2 in rng (the InternalRel of H1) by XTUPLE_0:def 13;
            hence contradiction by A16,A195,A191,A208,XBOOLE_0:3;
          end;
          [f.3,f.2] in (the InternalRel of H1) \/ (the InternalRel of H2
          ) by A138,A142,Def2;
          then
A212:     [f.3,f.2] in the InternalRel of H2 by A211,XBOOLE_0:def 3;
A213:     for x,y being Element of N4 holds [x,y] in the InternalRel of
          N4 iff [f.x,f.y] in the InternalRel of H2
          proof
            let x,y being Element of N4;
            thus [x,y] in the InternalRel of N4 implies [f.x,f.y] in the
            InternalRel of H2
            proof
              assume
A214:         [x,y] in the InternalRel of N4;
              per cases by A214,Th2,ENUMSET1:def 4;
              suppose
A215:           [x,y] = [0,1];
                then x = 0 by XTUPLE_0:1;
                hence thesis by A196,A215,XTUPLE_0:1;
              end;
              suppose
A216:           [x,y] = [1,0];
                then x = 1 by XTUPLE_0:1;
                hence thesis by A193,A216,XTUPLE_0:1;
              end;
              suppose
A217:           [x,y] = [1,2];
                then x = 1 by XTUPLE_0:1;
                hence thesis by A207,A217,XTUPLE_0:1;
              end;
              suppose
A218:           [x,y] = [2,1];
                then x = 2 by XTUPLE_0:1;
                hence thesis by A210,A218,XTUPLE_0:1;
              end;
              suppose
A219:           [x,y] = [2,3];
                then x = 2 by XTUPLE_0:1;
                hence thesis by A209,A219,XTUPLE_0:1;
              end;
              suppose
A220:           [x,y] = [3,2];
                then x = 3 by XTUPLE_0:1;
                hence thesis by A212,A220,XTUPLE_0:1;
              end;
            end;
            thus [f.x,f.y] in the InternalRel of H2 implies [x,y] in the
            InternalRel of N4
            proof
              [f.x,f.y] in the InternalRel of S implies [x,y] in the
              InternalRel of N4 by A137;
              then
A221:         [f.x,f.y] in (the InternalRel of H1) \/ (the InternalRel
              of H2) implies [x,y] in the InternalRel of N4 by A138,Def2;
              assume [f.x,f.y] in the InternalRel of H2;
              hence thesis by A221,XBOOLE_0:def 3;
            end;
          end;
A222:     f.3 in rng (the InternalRel of H2) by A209,XTUPLE_0:def 13;
          rng f c= the carrier of H2
          proof
            let y be object;
            assume y in rng f;
            then consider x being object such that
A223:       x in dom f and
A224:       y = f.x by FUNCT_1:def 3;
            per cases by A187,A223,Lm1,ENUMSET1:def 2;
            suppose
              x = 0;
              hence thesis by A186,A224;
            end;
            suppose
              x = 1;
              hence thesis by A195,A197,A224;
            end;
            suppose
              x = 2;
              hence thesis by A195,A208,A224;
            end;
            suppose
              x = 3;
              hence thesis by A195,A222,A224;
            end;
          end;
          then f is Function of the carrier of N4,the carrier of H2 by
FUNCT_2:6;
          then H2 embeds N4 by A136,A213,NECKLACE:def 1;
          hence thesis by A5,A18,A194,A204;
        end;
      end;
      hence thesis by A19,A20;
    end;
  end;
  assume R embeds N4;
  then P[k] by A1;
  then
A225: ex i being Nat st P[i];
  P[0] from NAT_1:sch 7(A225,A2);
  hence thesis;
end;
