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

theorem
  3 <= k implies primenumber(k) + primenumber(k+1) <= Product primesFinS k
  proof
    assume
A1: 3 <= k;
    set P = Product primesFinS k;
    P >= primenumber 0 * primenumber 1 * primenumber 2 by A1,Lm12;
    then P > 6 by XXREAL_0:2,MOEBIUS2:8,10,12;
    then consider a,b such that
A2: a > 1 and
A3: b > 1 and
A4: P = a+b and
A5: a,b are_coprime by NUMBER05:22;
    consider p,q being Prime such that
A6: p divides a and
A7: not p divides b and
A8: q divides b and
A9: not q divides a and
A10: p <> q by A2,A3,A5,Th8;
    p <= a & q <= b by A2,A3,A6,A8,NAT_D:7;
    then
A11: p+q <= a+b by XREAL_1:7;
    per cases by A10,XXREAL_0:1;
    suppose
A12:  p < q;
A13:  not p divides P
      proof
        assume p divides P;
        then p divides a+b-a by A4,A6,INT_5:1;
        hence contradiction by A7;
      end;
      now
        assume not p,P are_coprime;
        then consider z being Prime such that
A14:    z divides p and
A15:    z divides P by PYTHTRIP:def 2;
        z = 1 or z = p by A14,INT_2:def 4;
        hence contradiction by A13,A15,INT_2:def 4;
      end;
      then
A16:  primenumber(k) <= p by Th16;
      then per cases by XXREAL_0:1;
      suppose primenumber(k) < p;
        then primenumber(k+1) <= p by PRIMRECI:12;
        then primenumber(k+1) < q by A12,XXREAL_0:2;
        then primenumber(k) + primenumber(k+1) <= p+q by A16,XREAL_1:7;
        hence thesis by A4,A11,XXREAL_0:2;
      end;
      suppose primenumber(k) = p;
        then primenumber(k+1) <= q by A12,PRIMRECI:12;
        then primenumber(k) + primenumber(k+1) <= p+q by A16,XREAL_1:7;
        hence thesis by A4,A11,XXREAL_0:2;
      end;
    end;
    suppose
A17:  q < p;
A18:  not q divides P
      proof
        assume q divides P;
        then q divides a+b-b by A4,A8,INT_5:1;
        hence contradiction by A9;
      end;
      now
        assume not q,P are_coprime;
        then consider z being Prime such that
A19:    z divides q and
A20:    z divides P by PYTHTRIP:def 2;
        z = 1 or z = q by A19,INT_2:def 4;
        hence contradiction by A18,A20,INT_2:def 4;
      end;
      then
A21:  primenumber(k) <= q by Th16;
      then per cases by XXREAL_0:1;
      suppose primenumber(k) < q;
        then primenumber(k+1) <= q by PRIMRECI:12;
        then primenumber(k+1) < p by A17,XXREAL_0:2;
        then primenumber(k) + primenumber(k+1) <= p+q by A21,XREAL_1:7;
        hence thesis by A4,A11,XXREAL_0:2;
      end;
      suppose primenumber(k) = q;
        then primenumber(k+1) <= p by A17,PRIMRECI:12;
        then primenumber(k) + primenumber(k+1) <= p+q by A21,XREAL_1:7;
        hence thesis by A4,A11,XXREAL_0:2;
      end;
    end;
  end;
