theorem Th45:
  for a,b be Nat st a mod 2 = b mod 2 holds (-1)|^a = (-1)|^b
proof
  let a,b be Nat;
  assume a mod 2 = b mod 2;
  then a,b are_congruent_mod 2 by NAT_D:64;
  then
A1: 2 divides (a-b);
  per cases;
  suppose
    a>=b;
    then reconsider l=a-b as Element of NAT by NAT_1:21;
    consider n be Nat such that
A2: l=2*n by A1,NAT_D:def 3;
    (-1)|^a = (-1)|^(b + (2*n)) by A2
      .= ((-1)|^b) * ((-1)|^(2*n)) by NEWTON:8
      .= (-1)|^b * ((-1)|^2)|^n by NEWTON:9
      .= (-1)|^b * (1|^2)|^n by WSIERP_1:1
      .= (-1)|^b * 1;
    hence thesis;
  end;
  suppose a<b;
    then reconsider l=b-a as Element of NAT by NAT_1:21;
    2 divides -(a-b) by A1,INT_2:10;
    then consider n be Nat such that
A3: l=2*n by NAT_D:def 3;
    (-1)|^b = (-1)|^(a+2*n) by A3
      .= (-1)|^a * (-1)|^(2*n) by NEWTON:8
      .= (-1)|^a * ((-1)|^2)|^n by NEWTON:9
      .= (-1)|^a * (1|^2)|^n by WSIERP_1:1
      .= (-1)|^a * 1;
    hence thesis;
  end;
end;
