NO Prover = TRS(tech=GUIDED_UNF, nb_unfoldings=unlimited, unfold_variables=true, strategy=LEFTMOST_NE) ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 3] f(g(g(g(_0))),g(g(_0)),g(g(g(_0)))) -> f(g(g(g(_0))),g(g(_0)),g(g(g(_0)))) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {}. We have r|p = f(g(g(g(_0))),g(g(_0)),g(g(g(_0)))) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = f(g(g(g(_0))),g(g(_0)),g(g(g(_0)))) loops w.r.t. the analyzed TRS. ** END proof argument ** ** BEGIN proof description ** ## Searching for a generalized rewrite rule (a rule whose right-hand side contains a variable that does not occur in the left-hand side)... No generalized rewrite rule found! ## Searching for a loop by unfolding (unfolding of variable subterms: ON)... # Iteration 0: no loop detected, 1 unfolded rule generated. # Iteration 1: no loop detected, 3 unfolded rules generated. # Iteration 2: no loop detected, 8 unfolded rules generated. # Iteration 3: loop detected, 1 unfolded rule generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = f^#(g(_0),_0,_1) -> f^#(_1,_1,g(_1)) is in U_IR^0. Let p0 = [0]. The subterm at position p0 in the left-hand side of the rule of L0 unifies with the subterm at position p0 in the right-hand side of the rule of L0. ==> L1 = f^#(g(_0),_0,g(_0)) -> f^#(g(_0),g(_0),g(g(_0))) is in U_IR^1. Let p1 = [1]. We unfold the rule of L1 forwards at position p1 with the rule g(g(_0)) -> g(_0). ==> L2 = f^#(g(g(_0)),g(_0),g(g(_0))) -> f^#(g(g(_0)),g(_0),g(g(g(_0)))) is in U_IR^2. Let p2 = [2, 0, 0]. We unfold the rule of L2 forwards at position p2 with the rule g(g(_0)) -> g(_0). ==> L3 = f^#(g(g(g(_0))),g(g(_0)),g(g(g(_0)))) -> f^#(g(g(g(_0))),g(g(_0)),g(g(g(_0)))) is in U_IR^3. ** END proof description ** Proof stopped at iteration 3 Number of unfolded rules generated by this proof = 13 Number of unfolded rules generated by all the parallel proofs = 13