reserve X for set;
reserve a,b,c,k,m,n for Nat;
reserve i,j for Integer;
reserve r,s for Real;
reserve p,p1,p2,p3 for Prime;

theorem Th49:
  p1,p2,p3 are_mutually_distinct implies
  p1 >= 2 & p2 >= 3 & p3 >= 5 or p1 >= 2 & p2 >= 5 & p3 >= 3 or
  p1 >= 3 & p2 >= 2 & p3 >= 5 or p1 >= 3 & p2 >= 5 & p3 >= 2 or
  p1 >= 5 & p2 >= 2 & p3 >= 3 or p1 >= 5 & p2 >= 3 & p3 >= 2
  proof
    assume that
A1: p1,p2,p3 are_mutually_distinct and
A2: p1 < 2 or p2 < 3 or p3 < 5 and
A3: p1 < 2 or p2 < 5 or p3 < 3 and
A4: p1 < 3 or p2 < 2 or p3 < 5 and
A5: p1 < 3 or p2 < 5 or p3 < 2 and
A6: p1 < 5 or p2 < 2 or p3 < 3 and
A7: p1 < 5 or p2 < 3 or p3 < 2;
    per cases by A2;
    suppose p1 < 1+1;
      then p1 <= 1 by NAT_1:13;
      hence contradiction by INT_2:def 4;
    end;
    suppose that
      p1 >= 2 and
A8:   p2 < 3;
A9:  p2 = 2 by A8,NUMBER03:65;
      then p3 = 3 by A1,A4,XPRIMET1:3,NUMBER03:65;
      hence contradiction by A1,A6,A9,XPRIMET1:3;
    end;
    suppose that
A10:  p1 >= 2 and
A11:  p2 >= 3 and
A12:  p3 < 5;
      per cases by A12,XPRIMET1:3;
      suppose
A13:    p3 = 2;
        then p1 = 3 by A1,A7,A11,XPRIMET1:3;
        hence contradiction by A1,A5,A13,XPRIMET1:3;
      end;
      suppose p3 = 3;
        then p2 = 2 by A1,A3,A10,XPRIMET1:3;
        hence contradiction by A11;
      end;
    end;
  end;
