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% An element z is called nilpotent if z^n = 0 for some positive integer n. 
%  Show that 0 is the only nilpotent in a domain.
%  This is an ordinary Otter file in which the required instance of induction is given.
%  It shows that Otter can do inductive proofs if you give it the right instances of induction.

op(450, xfx, +).  % sum
op(350, fy, ~).   % minus
op(400, xfx, *).  % product
set(knuth_bendix).
set(hyper_res).
set(unit_deletion).
set(factor).
assign(pick_given_ratio,4).

list(sos).
x=x.
% ring axioms
 0 + x = x.
 x + 0 = x.
 ~x + x = 0.
 x + ~x = 0.
 (x + y) + z = x + (y + z).
 x + y = y + x.
(x * y) * z = x * (y * z).     % product is associative
 x * (y + z) = x * y + x * z.   % right distributivity
 (y + z) * x = y * x + z * x.   % left distributivity
 x * 1 = x.
 1 * x = x.
 x * y != 0  | x = 0 | y = 0.    %  no zero divisors  ("domain")
 pow(y+1,z) = z * pow(y,z).
 pow(1,z) = z.      
 pow(o,z) = 1.        % pow(o,0)  included.
end_of_list.

% P(x,y) is   pow(x,y) != 0.
% Induction(P) is   -P(o,y) | all(x,(P(x,y) & -P(x+1,y)) | P(z,y).
% Bringing induction to clausal form we have (using a Skolem function g)
% -P(o,u) | P(g(u,z),u) | P(z,u).   [1]
% -P(o,u) | -P(g(u,z)+1,u) | P(z,u). [2]

list(usable).
1 != 0.
pow(o,u) = 0 | pow(g(u,z),u) != 0 | pow(z,u) != 0.     % This is [1]
pow(o,u) = 0 | pow(g(u,z)+1,u) = 0 | pow(z,u) != 0.   % This is [2]
b != 0.
end_of_list.
 
list(sos).
pow(c,b) = 0.
end_of_list.