reserve a,b,i,k,m,n for Nat;
reserve s,z for non zero Nat;
reserve r for Real;
reserve c for Complex;
reserve e1,e2,e3,e4,e5 for ExtReal;

theorem Th62:
  5 divides n+1 or 5 divides n+7 or 5 divides n+9 or
  5 divides n+13 or 5 divides n+15
  proof
    assume that
A1: not 5 divides n+1 and
A2: not 5 divides n+7 and
A3: not 5 divides n+9 and
A4: not 5 divides n+13;
A5: 1 mod 5 = 1 by NAT_D:24;
A6: 2 mod 5 = 2 by NAT_D:24;
A7: 3 mod 5 = 3 by NAT_D:24;
A8: 4 mod 5 = 4 by NAT_D:24;
A9: 5*1+2 mod 5 = 2 mod 5 by NAT_D:21;
A10: 5*2+3 mod 5 = 3 mod 5 by NAT_D:21;
A11: 5*1+4 mod 5 = 4 mod 5 by NAT_D:21;
A12: 5*3 mod 5 = 0 by NAT_D:13;
A13: n+1-1 mod 5 = ((n+1 mod 5)-(1 mod 5)) mod 5 by INT_6:7;
A14: (n+15) mod 5 = ((n mod 5)+(15 mod 5)) mod 5 by NAT_D:66;
    per cases by A1,Lm18;
    suppose n+1 mod 5 = 1;
      hence 5 divides n+15 by A12,A13,A14,INT_1:62;
    end;
    suppose n+1 mod 5 = 2;
      then n+9 mod 5 = (2-1+4) mod 5 by A13,A5,A8,A11,NAT_D:66
      .= 0 by NAT_D:25;
      hence 5 divides n+15 by A3,INT_1:62;
    end;
    suppose (n+1) mod 5 = 3;
      then n+13 mod 5 = (3-1+3) mod 5 by A5,A6,A13,A10,NAT_D:66
      .= 0 by NAT_D:25;
      hence 5 divides n+15 by A4,INT_1:62;
    end;
    suppose (n+1) mod 5 = 4;
      then n+7 mod 5 = (4-1+2) mod 5 by A5,A9,A6,A7,A13,NAT_D:66
      .= 0 by NAT_D:25;
      hence 5 divides n+15 by A2,INT_1:62;
    end;
  end;
