reserve X for set;
reserve a,b,c,k,m,n for Nat;
reserve i,j for Integer;
reserve r for Real;
reserve p,p1,p2 for Prime;

theorem
  n > 17 implies n satisfies_Sierpinski_problem_48
  proof
    assume
A1: n > 17;
    per cases;
    suppose
A2:   n is even;
      n > 8 by A1,XXREAL_0:2;
      hence thesis by A2,Th41;
    end;
    suppose
A3:   n is odd;
      consider k such that
A4:   n = 12*k+(0 qua Nat) or ... or n = 12*k+(12-1) by NUMBER02:22;
      consider i being Nat such that
A5:   0 <= i <= 11 and
A6:   n = 12*k+i by A4;
A7:   i = 0 or ... or i = 11 by A5;
A8:  now
        assume 9 divides 6*k-1;
        then consider z being Integer such that
A9:     6*k-1 = 9*z;
        3 divides 3*(2*k-3*z);
        hence contradiction by A9,NAT_D:7;
      end;
A10:  now
        assume 9 divides 6*k+1;
        then consider z being Integer such that
A11:    6*k+1 = 9*z;
        3 divides 3*(3*z-2*k);
        hence contradiction by A11,NAT_D:7;
      end;
      12*k+0 = 2*(6*k) & 12*k+2 = 2*(6*k+1) & 12*k+4 = 2*(6*k+2) &
      12*k+6 = 2*(6*k+3) & 12*k+8 = 2*(6*k+4) & 12*k+10 = 2*(6*k+5);
      then per cases by A3,A6,A7;
      suppose
A12:     n = 12*k+1;
        then 12*k+1-1 > 17-1 by A1,XREAL_1:9;
        then 12*k/12 > 16/12 by XREAL_1:97;
        then
A13:    k > 1 by XXREAL_0:2;
        then
A14:    6*k > 6*1 by XREAL_1:68;
        k-1 > 1-1 by A13,XREAL_1:9;
        then k-1 >= 0+1 by INT_1:7;
        then 6*(k-1) >= 6*1 by XREAL_1:64;
        then
A15:    6*(k-1)-1 >= 6-1 by XREAL_1:9;
        then reconsider a = 6*(k-1)-1 as Element of NAT by INT_1:3;
A16:    6*k-1 > 6-1 by A14,XREAL_1:9;
        then reconsider b = 6*(k-1)+5 as Element of NAT by INT_1:3;
        set c = 9;
        take a, b, c;
        thus n = a+b+c by A12;
        thus a > 1 by A15,XXREAL_0:2;
        thus b > 1 by A16,XXREAL_0:2;
        thus c > 1;
A17:    1 divides a by INT_2:12;
A18:    1 divides b by INT_2:12;
A19:    1 divides c by INT_2:12;
        for d being Nat st d divides a & d divides b holds d divides 1
        proof
          let d be Nat such that
A20:      d divides a and
A21:      d divides b;
A22:      d divides b-a by A20,A21,INT_5:1;
A23:      3 divides 3*(2*(k-1)) & 6 divides 6*(k-1);
          d = 1 or d = 2 or d = 3 or d = 6 by A22,Th10;
          hence thesis by A20,A23,Th4;
        end;
        hence a,b are_coprime by A17,A18,NAT_D:def 5;
        for d being Nat st d divides a & d divides c holds d divides 1
        proof
          let d be Nat such that
A24:      d divides a and
A25:      d divides c;
A26:      3 divides 3*(2*(k-1));
A27:      now
            assume 9 divides 6*k-7;
            then consider z being Integer such that
A28:        6*k-7 = 9*z;
A29:        3 divides 3*(2*k-3*z);
            (2*3+1) mod 3 = 1 mod 3 by NAT_D:21
            .= 1 by NAT_D:24;
            hence contradiction by A28,A29,INT_1:62;
          end;
          d = 1 or d = 3 or d = 9 by A25,Th11;
          hence thesis by A24,A26,A27,Th4;
        end;
        hence a,c are_coprime by A17,A19,NAT_D:def 5;
        for d being Nat st d divides b & d divides c holds d divides 1
        proof
          let d be Nat such that
A30:      d divides b and
A31:      d divides c;
A32:      3 divides 3*(2*k);
          d = 1 or d = 3 or d = 9 by A31,Th11;
          hence thesis by A30,A32,A8,Th4;
        end;
        hence b,c are_coprime by A18,A19,NAT_D:def 5;
      end;
      suppose
A33:     n = 12*k+3;
        then 12*k+3-3 > 17-3 by A1,XREAL_1:9;
        then 12*k/12 > (1*12+2)/12 by XREAL_1:97;
        then k > 1 by XXREAL_0:2;
        then
A34:    6*k > 6*1 by XREAL_1:68;
        then
A35:    6*k-1 > 6-1 by XREAL_1:9;
        then reconsider a = 6*k-1 as Element of NAT by INT_1:3;
        set b = 6*k+1;
        set c = 3;
        take a, b, c;
        thus n = a+b+c by A33;
        thus a > 1 by A35,XXREAL_0:2;
        6*k+1 > 6+1 by A34,XREAL_1:6;
        hence b > 1 by XXREAL_0:2;
        thus c > 1;
A36:    1 divides a & 1 divides b by INT_2:12;
        for d being Nat st d divides a & d divides b holds d divides 1
        proof
          let d be Nat such that
A37:      d divides a and
A38:      d divides b;
A39:      a = 2*(3*k)-1;
          d divides b-a by A37,A38,INT_5:1;
          then d = 1 or d = 2 by Th9;
          hence thesis by A37,A39;
        end;
        hence a,b are_coprime by A36,NAT_D:def 5;
A40:    3 divides 3*(2*k);
        hence a,c are_coprime by Th6;
        thus b,c are_coprime by A40,Th5;
      end;
      suppose
A41:     n = 12*k+5;
        then 12*k+5-5 > 17-5 by A1,XREAL_1:9;
        then 12*k/12 > 12/12 by XREAL_1:97;
        then
A42:    6*k > 6*1 by XREAL_1:68;
        then
A43:    6*k-5 > 6-5 by XREAL_1:9;
        then reconsider a = 6*k-5 as Element of NAT by INT_1:3,XXREAL_0:2;
        set b = 6*k+1;
        set c = 9;
        take a, b, c;
        thus n = a+b+c by A41;
        thus a > 1 by A43;
        6*k+1 > 6+1 by A42,XREAL_1:8;
        hence b > 1 by XXREAL_0:2;
        thus c > 1;
A44:    1 divides a by INT_2:12;
A45:    1 divides b by INT_2:12;
A46:    1 divides c by INT_2:12;
        for d being Nat st d divides a & d divides b holds d divides 1
        proof
          let d be Nat such that
A47:      d divides a and
A48:      d divides b;
A49:      3 divides 3*(2*k) & 6 divides 6*k;
          d divides b-a by A47,A48,INT_5:1;
          then d = 1 or d = 2 or d = 3 or d = 6 by Th10;
          hence thesis by A48,A49,Th3;
        end;
        hence a,b are_coprime by A44,A45,NAT_D:def 5;
        for d being Nat st d divides a & d divides c holds d divides 1
        proof
          let d be Nat such that
A50:      d divides a and
A51:      d divides c;
A52:      3 divides 3*(2*(k-1));
A53:      6*k-5 = 6*(k-1)+1;
A54:      now
            assume 9 divides 6*k-5;
            then consider z being Integer such that
A55:        6*k-5 = 9*z;
A56:        3 divides 3*(2*k-3*z);
            (1*3+2) mod 3 = 2 mod 3 by NAT_D:21
            .= 2 by NAT_D:24;
            hence contradiction by A55,A56,INT_1:62;
          end;
          d = 1 or d = 3 or d = 9 by A51,Th11;
          hence thesis by A50,A52,A53,A54,Th3;
        end;
        hence a,c are_coprime by A44,A46,NAT_D:def 5;
        for d being Nat st d divides b & d divides c holds d divides 1
        proof
          let d be Nat such that
A57:      d divides b and
A58:      d divides c;
A59:      3 divides 3*(2*k);
          d = 1 or d = 3 or d = 9 by A58,Th11;
          hence thesis by A57,A59,A10,Th3;
        end;
        hence b,c are_coprime by A45,A46,NAT_D:def 5;
      end;
      suppose
A60:    n = 12*k+7;
        set a = 6*k+6-1;
        12*k+7-7 > 17-7 by A1,A60,XREAL_1:9;
        then 12*k/12 > 10/12 by XREAL_1:97;
        then
A61:    k >= 0+1 by NAT_1:13;
        then reconsider b = 6*k-1 as Element of NAT by INT_1:3;
        set c = 3;
        take a, b, c;
        thus n = a+b+c by A60;
A62:    6*k >= 6*1 by A61,XREAL_1:64;
        then 6*k+5 >= 6+5 by XREAL_1:6;
        hence a > 1 by XXREAL_0:2;
        6*k-1 >= 6-1 by A62,XREAL_1:9;
        hence b > 1 by XXREAL_0:2;
        thus c > 1;
A63:    1 divides a by INT_2:12;
A64:    1 divides b by INT_2:12;
        for d being Nat st d divides a & d divides b holds d divides 1
        proof
          let d be Nat such that
A65:      d divides a and
A66:      d divides b;
A67:      3 divides 3*(2*k) & 6 divides 6*k;
          d divides a-b by A65,A66,INT_5:1;
          then d = 1 or d = 2 or d = 3 or d = 6 by Th10;
          hence thesis by A66,A67,Th4;
        end;
        hence a,b are_coprime by A63,A64,NAT_D:def 5;
        3 divides 3*(2*(k+1));
        hence a,c are_coprime by Th6;
        3 divides 3*(2*k);
        hence b,c are_coprime by Th6;
      end;
      suppose
A68:    n = 12*k+9;
        then 12*k+9-9 > 17-9 by A1,XREAL_1:9;
        then
A69:    12*k/12 > 8/12 by XREAL_1:97;
        then reconsider a = 6*k-1 as Element of NAT by INT_1:3;
        set b = 6*k+1;
        set c = 9;
        take a, b, c;
        thus n = a+b+c by A68;
        k >= 0+1 by A69,NAT_1:13;
        then
A70:    6*k >= 6*1 by XREAL_1:64;
        then 6*k-1 >= 6-1 by XREAL_1:9;
        hence a > 1 by XXREAL_0:2;
        6*k+1 >= 6+1 by A70,XREAL_1:6;
        hence b > 1 by XXREAL_0:2;
        thus c > 1;
A71:    1 divides a by INT_2:12;
A72:    1 divides b by INT_2:12;
A73:    1 divides c by INT_2:12;
        for d being Nat st d divides a & d divides b holds d divides 1
        proof
          let d be Nat such that
A74:      d divides a and
A75:      d divides b;
A76:      a = 2*(3*k)-1;
          d divides b-a by A74,A75,INT_5:1;
          then d = 1 or d = 2 by Th9;
          hence thesis by A74,A76;
        end;
        hence a,b are_coprime by A71,A72,NAT_D:def 5;
        for d being Nat st d divides a & d divides c holds d divides 1
        proof
          let d be Nat such that
A77:      d divides a and
A78:      d divides c;
A79:      3 divides 3*(2*k);
          d = 1 or d = 3 or d = 9 by A78,Th11;
          hence thesis by A8,A77,A79,Th4;
        end;
        hence a,c are_coprime by A71,A73,NAT_D:def 5;
        for d being Nat st d divides b & d divides c holds d divides 1
        proof
          let d be Nat such that
A80:      d divides b and
A81:      d divides c;
A82:      3 divides 3*(2*k);
          d = 1 or d = 3 or d = 9 by A81,Th11;
          hence thesis by A10,A80,A82,Th3;
        end;
        hence b,c are_coprime by A72,A73,NAT_D:def 5;
      end;
      suppose
A83:    n = 12*k+11;
        set a = 6*k+1;
        set b = 6*k+6+1;
        set c = 3;
        take a, b, c;
        thus n = a+b+c by A83;
        12*k+11-11 > 17-11 by A1,A83,XREAL_1:9;
        then 12*k/12 > 6/12 by XREAL_1:97;
        then k >= 0+1 by NAT_1:13;
        then
A84:    6*k >= 6*1 by XREAL_1:64;
        then 6*k+1 >= 6+1 by XREAL_1:6;
        hence a > 1 by XXREAL_0:2;
        6*k+7 >= 6+7 by A84,XREAL_1:6;
        hence b > 1 by XXREAL_0:2;
        thus c > 1;
A85:    1 divides a by INT_2:12;
A86:    1 divides b by INT_2:12;
        for d being Nat st d divides a & d divides b holds d divides 1
        proof
          let d be Nat such that
A87:      d divides a and
A88:      d divides b;
A89:      3 divides 3*(2*k) & 6 divides 6*k;
          d divides b-a by A87,A88,INT_5:1;
          then d = 1 or d = 2 or d = 3 or d = 6 by Th10;
          hence thesis by A87,A89,Th3;
        end;
        hence a,b are_coprime by A85,A86,NAT_D:def 5;
        3 divides 3*(2*k);
        hence a,c are_coprime by Th5;
        3 divides 3*(2*k+2);
        hence b,c are_coprime by Th5;
      end;
    end;
  end;
