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

theorem Th63:
  for k being Nat holds (FIB | (OddNAT /\ Seg (2 * k + 3))) \/ {[2
  *k+5,FIB.(2 * k + 5)]} = (FIB | (OddNAT /\ Seg (2 * k + 5)))
proof
  let k be Nat;
A1: dom FIB = NAT by FUNCT_2:def 1;
  (FIB | (OddNAT /\ Seg (2 * k + 5))) = (FIB | (OddNAT /\ Seg (2 * k + 3)
  \/ {2 * k + 5})) by Th62
    .= FIB | (OddNAT /\ Seg (2 * k + 3)) \/ (FIB | ({2 * k + 5})) by RELAT_1:78
    .= FIB | (OddNAT /\ Seg (2 * k + 3)) \/ {[2*k+5,FIB.(2 * k + 5)]} by A1,
GRFUNC_1:28;
  hence thesis;
end;
