NO Prover = TRS(tech=GUIDED_UNF_TRIPLES, nb_unfoldings=unlimited, unfold_variables=false, max_nb_coefficients=12, max_nb_unfolded_rules=-1, strategy=LEFTMOST_NE) ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 5] f(_0,c(_0),c(g(_1,c(_0)))) -> f(_1,c(_0),c(g(_1,c(_0)))) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {_0->_1} and theta2 = {}. We have r|p = f(_1,c(_0),c(g(_1,c(_0)))) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = f(_1,c(_1),c(g(_1,c(_1)))) 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! ## Applying the DP framework... ## Round 1: ## DP problem: Dependency pairs = [f^#(_0,c(_0),c(_1)) -> f^#(_1,_1,f(_1,_0,_1)), f^#(_0,c(_0),c(_1)) -> f^#(_1,_0,_1)] TRS = {f(_0,c(_0),c(_1)) -> f(_1,_1,f(_1,_0,_1)), f(s(_0),_1,_2) -> f(_0,s(c(_1)),c(_2)), f(c(_0),_0,_1) -> c(_1), g(_0,_1) -> _0, g(_0,_1) -> _1} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Too many coefficients (17)! Aborting! ## Trying with lexicographic path orders... Failed! ## Trying to prove nontermination by unfolding the dependency pairs with the rules of the TRS # max_depth=3, unfold_variables=false: # Iteration 0: nontermination not detected, 2 unfolded rules generated. # Iteration 1: nontermination not detected, 4 unfolded rules generated. # Iteration 2: nontermination not detected, 4 unfolded rules generated. # Iteration 3: nontermination not detected, 2 unfolded rules generated. # Iteration 4: nontermination not detected, 2 unfolded rules generated. # Iteration 5: nontermination not detected, 0 unfolded rule generated. Nontermination not detected! # max_depth=3, unfold_variables=true: # Iteration 0: nontermination not detected, 2 unfolded rules generated. # Iteration 1: nontermination not detected, 4 unfolded rules generated. # Iteration 2: nontermination not detected, 6 unfolded rules generated. # Iteration 3: nontermination not detected, 10 unfolded rules generated. # Iteration 4: nontermination not detected, 18 unfolded rules generated. # Iteration 5: nontermination detected, 11 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = f^#(_0,c(_0),c(_1)) -> f^#(_1,_1,f(_1,_0,_1)) [trans] is in U_IR^0. We build a unit triple from L0. ==> L1 = f^#(_0,c(_0),c(_1)) -> f^#(_1,_1,f(_1,_0,_1)) [unit] is in U_IR^1. Let p1 = [0]. We unfold the rule of L1 forwards at position p1 with the rule g(_0,_1) -> _0. ==> L2 = f^#(_0,c(_0),c(g(_1,_2))) -> f^#(_1,g(_1,_2),f(g(_1,_2),_0,g(_1,_2))) [unit] is in U_IR^2. Let p2 = [1]. We unfold the rule of L2 forwards at position p2 with the rule g(_0,_1) -> _1. ==> L3 = f^#(_0,c(_0),c(g(_1,_2))) -> f^#(_1,_2,f(g(_1,_2),_0,g(_1,_2))) [unit] is in U_IR^3. Let p3 = [2, 0]. We unfold the rule of L3 forwards at position p3 with the rule g(_0,_1) -> _1. ==> L4 = f^#(_0,c(_0),c(g(_1,_2))) -> f^#(_1,_2,f(_2,_0,g(_1,_2))) [unit] is in U_IR^4. Let p4 = [2]. We unfold the rule of L4 forwards at position p4 with the rule f(c(_0),_0,_1) -> c(_1). ==> L5 = f^#(_0,c(_0),c(g(_1,c(_0)))) -> f^#(_1,c(_0),c(g(_1,c(_0)))) [unit] is in U_IR^5. This DP problem is infinite. ** END proof description ** Proof stopped at iteration 5 Number of unfolded rules generated by this proof = 65 Number of unfolded rules generated by all the parallel proofs = 303