
theorem
  for m be Element of NAT holds ALGO_BPOW(0,0,m) = 1
  proof
    let m be Element of NAT;
    consider A,B be sequence of NAT such that
    ASC:
    ALGO_BPOW(0,0,m) = B. (LenBSeq 0) &
    A.0 = 0 mod m & B.0 = 1 &
    ( for i be Nat holds
    A.(i+1) = (A.i)*(A.i) mod m &
    B.(i+1) = BinBranch((B.i),(B.i)*(A.i) mod m,(Nat2BL.0).(i+1))) by Def1;
    ALGO_BPOW(0,0,m) = B.1 by ASC,BINARI_6:def 1
    .= BinBranch((B.0),(B.0)*(A.0) mod m,(Nat2BL.0).(0+1)) by ASC
    .= 1 by ASC,defBB,zerobs;
    hence thesis;
  end;
