reserve D for non empty set;
reserve s for FinSequence of D;
reserve m,n for Element of NAT;

theorem
  1 <= len s implies Op-Shift(s,0)= s
  proof
    assume A1:1 <= len s;
    then
    A2:len (Op-Shift(s,0)) = len s &
    for i be Nat st i in Seg (len s) holds (Op-Shift(s,0)).i
    = s.( (i-1+0) mod (len s) + 1) by Def3;
    A3: now let i be Nat;
    assume i in dom (Op-Shift(s,0));
    then
    A4:i in Seg (len s) by A2,FINSEQ_1:def 3;
    then
    A5: 1<=i & i <=len s by FINSEQ_1:1;
    then
    A6: 1-1 <= i-1 by XREAL_1:9;
    i < len s + 1 by A5,NAT_1:13;
    then A7:
    i -1 < len s + 1 -1 by XREAL_1:14;
    thus (Op-Shift(s,0)).i = s.((i-1+0) mod (len s) + 1) by A4,Def3,A1
    .= s.((i-1) + 1) by A7,A6,NAT_D:63
    .= s.i;
  end;
  thus thesis by A2,A3,FINSEQ_2:9;
end;
