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
  f is non empty implies f/.1 = (Rev f)/.len f & f/.len f = (Rev f)/.1
proof
A1: dom f = dom Rev f by Th57;
  assume
A2: f is non empty;
  then
A3: len f in dom f by Th6;
  1 in dom f by A2,Th6;
  hence f/.1 = f.1 by PARTFUN1:def 6
    .= (Rev f).len f by Th62
    .= (Rev f)/.len f by A3,A1,PARTFUN1:def 6;
  thus f/.len f = f.(len f) by A3,PARTFUN1:def 6
    .= (Rev f).1 by Th62
    .= (Rev f)/.1 by A2,A1,Th6,PARTFUN1:def 6;
end;
