
theorem Th6:
  for S being non empty RelStr, a, b being Element of S,
  i, j being Element of Net-Str (a,b),
  i9, j9 being Element of NAT st i9 = i & j9 = i9 + 1 & j9 = j holds
  (Net-Str (a, b).i = a implies Net-Str (a, b).j = b) &
  (Net-Str (a, b).i = b implies Net-Str (a, b).j = a)
proof
  let S be non empty RelStr;
  let a, b be Element of S,
  i, j be Element of Net-Str (a,b), i9, j9 be Element of NAT;
  assume that
A1: i9 = i and
A2: j9 = i9 + 1 and
A3: j9 = j;
  per cases;
  suppose
A4: a <> b;
    defpred C[Element of NAT] means ex k be Element of NAT st $1 = 2*k;
    thus Net-Str (a, b).i = a implies Net-Str (a, b).j = b
    proof
      assume
A5:   Net-Str (a, b).i = a;
      C[i9]
      proof
        assume
A6:     not C[i9];
        Net-Str (a, b).i = (a,b),....i by Def3
          .= b by A1,A6,Def1;
        hence thesis by A4,A5;
      end;
      then
A7:   for k be Element of NAT holds j9 <> 2*k by A2;
      Net-Str (a, b).j = (a,b),....j by Def3
        .= b by A3,A7,Def1;
      hence thesis;
    end;
    assume
A8: Net-Str (a, b).i = b;
A9: not C[i9]
    proof
      assume
A10:  C[i9];
      Net-Str (a, b).i = (a,b),....i by Def3
        .= a by A1,A10,Def1;
      hence thesis by A4,A8;
    end;
A11: C[j9]
    proof
      assume not C[j9];
      then ex kl be Element of NAT st ( j9 = 2*kl + 1) by SCHEME1:1;
      hence thesis by A2,A9;
    end;
    Net-Str (a, b).j = (a,b),....j by Def3
      .= a by A3,A11,Def1;
    hence thesis;
  end;
  suppose a = b;
    hence thesis by Th5;
  end;
end;
