reserve a,b,c,h for Integer;
reserve k,m,n for Nat;
reserve i,j,z for Integer;
reserve p for Prime;

theorem Th84:
  n >= 8 implies ex m being Multiple of 3 st m in seq(k,n) & m is odd
  proof
    assume
A1: n >= 8;
    per cases by NUMBER06:18;
    suppose 3 divides k;
      then
A2:   k is Multiple of 3 by Def15;
      per cases;
      suppose
A3:     k is even;
        reconsider m = k+1*(2+1) as Multiple of 3 by A2,Th58;
        take m;
A4:     1+k <= k+1+2 by NAT_1:12;
        3 <= n by A1,XXREAL_0:2;
        then k+3 <= k+n by XREAL_1:6;
        hence m in seq(k,n) by A4;
        m = k+2+1;
        hence thesis by A3;
      end;
      suppose
A5:     k is odd;
        reconsider m = k+2*(2+1) as Multiple of 3 by A2,Th58;
        take m;
A6:     1+k <= k+1+5 by NAT_1:12;
        6 <= n by A1,XXREAL_0:2;
        then k+6 <= k+n by XREAL_1:6;
        hence m in seq(k,n) by A6;
        thus thesis by A5;
      end;
    end;
    suppose 3 divides k+1;
      then
A7:   k+1 is Multiple of 3 by Def15;
      per cases;
      suppose
A8:     k is even;
        reconsider m = k+1+2*(2+1) as Multiple of 3 by A7,Th58;
        take m;
A9:     1+k <= k+1+6 by NAT_1:12;
        7 <= n by A1,XXREAL_0:2;
        then k+7 <= k+n by XREAL_1:6;
        hence m in seq(k,n) by A9;
        thus thesis by A8;
      end;
      suppose
A10:    k is odd;
        reconsider m = k+1+1*(2+1) as Multiple of 3 by A7,Th58;
        take m;
A11:    1+k <= k+1+3 by NAT_1:12;
        4 <= n by A1,XXREAL_0:2;
        then k+4 <= k+n by XREAL_1:6;
        hence m in seq(k,n) by A11;
        m = k+2*2;
        hence thesis by A10;
      end;
    end;
    suppose 3 divides k+2;
      then
A12:  k+2 is Multiple of 3 by Def15;
      per cases;
      suppose
A13:    k is even;
        reconsider m = k+2+1*(2+1) as Multiple of 3 by A12,Th58;
        take m;
A14:    1+k <= k+1+4 by NAT_1:12;
        5 <= n by A1,XXREAL_0:2;
        then k+5 <= k+n by XREAL_1:6;
        hence m in seq(k,n) by A14;
        m = k+2*2+1;
        hence thesis by A13;
      end;
      suppose
A15:    k is odd;
        reconsider m = k+2+2*(2+1) as Multiple of 3 by A12,Th58;
        take m;
A16:    1+k <= k+1+7 by NAT_1:12;
        k+8 <= k+n by A1,XREAL_1:6;
        hence m in seq(k,n) by A16;
        thus thesis by A15;
      end;
    end;
  end;
