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 Th41:
  n is even & n > 8 implies n satisfies_Sierpinski_problem_48
  proof
    assume that
A1: n is even and
A2: n > 8;
    consider k such that
A3: n = 6*k or n = 6*k+1 or n = 6*k+2 or n = 6*k+3 or
    n = 6*k+4 or n = 6*k+5 by NUMBER02:26;
    6*k+1 = 2*(3*k)+1 & 6*k+3 = 2*(3*k+1)+1 & 6*k+5 = 2*(3*k+2)+1;
    then per cases by A1,A3;
    suppose
A4:   n = 6*k;
      then
A5:   6*k-5 > 8-5 by A2,XREAL_1:9;
      then 6*k-5 > 3;
      then reconsider c = 6*(k-1)+1 as Element of NAT by INT_1:3;
      take a = 2, b = 3, c;
      thus n = a+b+c by A4;
      thus a > 1;
      thus b > 1;
      thus c > 1 by A5,XXREAL_0:2;
      thus a,b are_coprime by XPRIMES1:2,3,INT_2:30;
      2 divides 2*(3*(k-1));
      hence a,c are_coprime by Th5;
      3 divides 3*(2*(k-1));
      hence b,c are_coprime by Th5;
    end;
    suppose
A6:   n = 6*k+2;
      then
A7:   6*k+2-7 > 8-7 by A2,XREAL_1:9;
      then reconsider c = 6*(k-1)+1 as Element of NAT by INT_1:3;
      take a = 3, b = 4, c;
      thus n = a+b+c by A6;
      thus a > 1;
      thus b > 1;
      thus c > 1 by A7;
      4 = 3+1;
      hence a,b are_coprime by Th5;
      3 divides 3*(2*(k-1));
      hence a,c are_coprime by Th5;
A8:  1 divides b by INT_2:12;
A9:  1 divides c by INT_2:12;
      for d being Nat st d divides b & d divides c holds d divides 1
      proof
        let d be Nat such that
A10:    d divides b and
A11:    d divides c;
A12:    4 = 2*2;
A13:    c = 2*(3*(k-1))+1;
        d = 1 or d = 2 or d = 4 by A10,NUMBER05:21;
        hence thesis by A11,A12,A13;
      end;
      hence b,c are_coprime by A8,A9,NAT_D:def 5;
    end;
    suppose
A14:   n = 6*k+4;
      then
A15:   6*k+4-5 > 8-5 by A2,XREAL_1:9;
      then 6*k-1 > 3;
      then reconsider c = 6*k-1 as Element of NAT by INT_1:3;
      take a = 2, b = 3, c;
      thus n = a+b+c by A14;
      thus a > 1;
      thus b > 1;
      thus c > 1 by A15,XXREAL_0:2;
      thus a,b are_coprime by XPRIMES1:2,3,INT_2:30;
      2 divides 2*(3*k);
      hence a,c are_coprime by Th6;
      3 divides 3*(2*k);
      hence b,c are_coprime by Th6;
    end;
  end;
