reserve i,j,k,n for Nat;
reserve D for non empty set,
  p for Element of D,
  f,g for FinSequence of D;

theorem Th49:
  p in rng f implies ex i being Element of NAT st i+1 = p..f & f:- p = f/^i
proof
  assume
A1: p in rng f;
  then reconsider i = p..f - 1 as Element of NAT by FINSEQ_4:21,INT_1:5;
A2: p..f in dom f by A1,FINSEQ_4:20;
  take i;
  thus
A3: i+1 = p..f;
  thus f:-p = <*f/.(p..f)*>^(f/^p..f) by A1,Th38
    .= f/^i by A3,A2,Th31;
end;
