
theorem :: As above, similarly for difference = 1000
  not ex f being Arithmetic_Progression st
    difference f = 1000 &
      ex p1,p2,p3 being Prime, i being Nat st
      p1,p2,p3 are_mutually_distinct &
        p1 = f.i & p2 = f.(i+1) & p3 = f.(i+2)
  proof
    given f being Arithmetic_Progression such that
A1: difference f = 1000 & ex p1,p2,p3 being Prime, i be Nat st
      p1,p2,p3 are_mutually_distinct &
        p1 = f.i & p2 = f.(i+1) & p3 = f.(i+2);
    reconsider f as integer-valued Arithmetic_Progression by A1,LemmaIntProg;
    consider p1,p2,p3 being Prime, i be Nat such that
A2: p1,p2,p3 are_mutually_distinct &
        p1 = f.i & p2 = f.(i+1) & p3 = f.(i+2) by A1;
b1: p2 - p1 = difference f by A2,LemmaDiffConst;
    p3 = f.(i+1+1) by A2; then
b2: p3 - p2 = difference f by A2,LemmaDiffConst;
K1: 3 divides p1+1 iff 3 divides p1+1+3*333 by LemmaCong;
    3 divides p1+2 iff 3 divides p1+2+3*666 by LemmaCong; then
    per cases by LemmaDivides,b2,A1,b1,K1,ThreeConsecutive;
    suppose
AA:   p1 = 3;
      1003 = 17 * 59; then

::: ERROR in Sierpinski 250: 17 * 53 = 1003 which is simple typo

      17 divides 1003; then
      1003 is not prime;
      hence thesis by AA,b1,A1;
    end;
    suppose
      p2 = 3; then
      p1 = 3 - 1000 by b1,A1; then
      p1 < 0;
      hence thesis;
    end;
    suppose
      p3 = 3; then
      p2 = 3 - 1000 by b2,A1; then
      p2 < 0;
      hence thesis;
    end;
  end;
