/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 4] f(_0,i(_0)) -> f(i(_0),i(i(_0))) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {_0->i(_0)}. We have r|p = f(i(_0),i(i(_0))) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = f(_0,i(_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! ## Applying the DP framework... ## 1 initial DP problem to solve. ## First, we try to decompose this problem into smaller problems. ## Round 1 [1 DP problem]: ## DP problem: Dependency pairs = [f^#(_0,_0) -> f^#(i(_0),g(g(_0))), f^#(_0,i(_0)) -> f^#(_0,_0)] TRS = {f(_0,_0) -> f(i(_0),g(g(_0))), f(_0,_1) -> _0, g(_0) -> i(_0), f(_0,i(_0)) -> f(_0,_0), f(i(_0),i(g(_0))) -> a} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Failed! ## Trying with lexicographic path orders... Failed! ## Trying with Knuth-Bendix orders... Failed! Don't know whether this DP problem is finite. ## A DP problem could not be proved finite. ## Now, we try to prove that this problem is infinite. ## Trying to find a loop (forward=true, backward=true, max=20) # max_depth=20, unfold_variables=false: # Iteration 0: no loop found, 2 unfolded rules generated. # Iteration 1: no loop found, 4 unfolded rules generated. # Iteration 2: no loop found, 4 unfolded rules generated. # Iteration 3: no loop found, 6 unfolded rules generated. # Iteration 4: success, found a loop, 1 unfolded rule generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = f^#(_0,i(_0)) -> f^#(_0,_0) [trans] is in U_IR^0. D = f^#(_0,_0) -> f^#(i(_0),g(g(_0))) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = f^#(_0,i(_0)) -> f^#(i(_0),g(g(_0))) [trans] is in U_IR^1. We build a unit triple from L1. ==> L2 = f^#(_0,i(_0)) -> f^#(i(_0),g(g(_0))) [unit] is in U_IR^2. Let p2 = [1]. We unfold the rule of L2 forwards at position p2 with the rule g(_0) -> i(_0). ==> L3 = f^#(_0,i(_0)) -> f^#(i(_0),i(g(_0))) [unit] is in U_IR^3. Let p3 = [1, 0]. We unfold the rule of L3 forwards at position p3 with the rule g(_0) -> i(_0). ==> L4 = f^#(_0,i(_0)) -> f^#(i(_0),i(i(_0))) [unit] is in U_IR^4. This DP problem is infinite. Proof run on Linux version 3.10.0-1160.25.1.el7.x86_64 for amd64 using Java version 1.8.0_292 ** END proof description ** Total number of generated unfolded rules = 66