reserve n, k, r, m, i, j for Nat;

theorem Th62:
  for k being Nat holds OddNAT /\ Seg (2 * k + 3) \/ {2 * k + 5} =
  OddNAT /\ Seg (2 * k + 5)
proof
  let k be Nat;
  2 * k + 5 = 2 * (k + 2) + 1;
  then
A1: 2 * k + 5 in OddNAT;
  2 * k + 4 = 2 * ((k+1) +1);
  then
A2: {2 * k + 4} misses OddNAT by Th52,ZFMISC_1:50;
  OddNAT /\ Seg (2 * k + 5) = OddNAT /\ Seg (2 * k + 4 + 1)
    .= OddNAT /\ (Seg (2 * k + 4) \/ {2 * k + 5}) by FINSEQ_1:9
    .= OddNAT /\ Seg (2 * k + 3 + 1) \/ OddNAT /\ {2 * k + 5} by XBOOLE_1:23
    .= OddNAT /\ (Seg (2 * k + 3) \/ {2 * k + 4}) \/ OddNAT /\ {2 * k + 5}
  by FINSEQ_1:9
    .= OddNAT /\ Seg (2 * k + 3) \/ OddNAT /\ {2 * k + 4} \/ OddNAT /\ {2 *
  k + 5} by XBOOLE_1:23
    .= OddNAT /\ Seg (2 * k + 3) \/ {} \/ OddNAT /\ {2 * k + 5} by A2
    .= OddNAT /\ Seg (2 * k + 3) \/ {2 * k + 5} by A1,ZFMISC_1:46;
  hence thesis;
end;
