/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 6] a__f(_0,_0) -> a__f(b,b) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {_0->b}. We have r|p = a__f(b,b) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = a__f(_0,_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 = [a__h^#(_0) -> a__g^#(mark(_0),_0), a__f^#(_0,_0) -> a__h^#(a__a), mark^#(f(_0,_1)) -> a__f^#(mark(_0),_1), a__g^#(a,_0) -> a__f^#(b,_0), mark^#(g(_0,_1)) -> a__g^#(mark(_0),_1), a__h^#(_0) -> mark^#(_0), mark^#(h(_0)) -> a__h^#(mark(_0)), mark^#(h(_0)) -> mark^#(_0), mark^#(g(_0,_1)) -> mark^#(_0), mark^#(f(_0,_1)) -> mark^#(_0)] TRS = {a__h(_0) -> a__g(mark(_0),_0), a__g(a,_0) -> a__f(b,_0), a__f(_0,_0) -> a__h(a__a), a__a -> b, mark(h(_0)) -> a__h(mark(_0)), mark(g(_0,_1)) -> a__g(mark(_0),_1), mark(a) -> a__a, mark(f(_0,_1)) -> a__f(mark(_0),_1), mark(b) -> b, a__h(_0) -> h(_0), a__g(_0,_1) -> g(_0,_1), a__a -> a, a__f(_0,_1) -> f(_0,_1)} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... This DP problem is too complex! Aborting! ## 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=false, max=-1) # max_depth=2, unfold_variables=false: # Iteration 0: no loop found, 10 unfolded rules generated. # Iteration 1: no loop found, 39 unfolded rules generated. # Iteration 2: no loop found, 62 unfolded rules generated. # Iteration 3: no loop found, 80 unfolded rules generated. # Iteration 4: no loop found, 89 unfolded rules generated. # Iteration 5: no loop found, 97 unfolded rules generated. # Iteration 6: success, found a loop, 5 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = a__f^#(_0,_0) -> a__h^#(a__a) [trans] is in U_IR^0. D = a__h^#(_0) -> a__g^#(mark(_0),_0) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = a__f^#(_0,_0) -> a__g^#(mark(a__a),a__a) [trans] is in U_IR^1. D = a__g^#(a,_0) -> a__f^#(b,_0) is a dependency pair of IR. We build a composed triple from L1 and D. ==> L2 = [a__f^#(_0,_0) -> a__g^#(mark(a__a),a__a), a__g^#(a,_1) -> a__f^#(b,_1)] [comp] is in U_IR^2. Let p2 = [0, 0]. We unfold the first rule of L2 forwards at position p2 with the rule a__a -> a. ==> L3 = [a__f^#(_0,_0) -> a__g^#(mark(a),a__a), a__g^#(a,_1) -> a__f^#(b,_1)] [comp] is in U_IR^3. Let p3 = [0]. We unfold the first rule of L3 forwards at position p3 with the rule mark(a) -> a__a. ==> L4 = [a__f^#(_0,_0) -> a__g^#(a__a,a__a), a__g^#(a,_1) -> a__f^#(b,_1)] [comp] is in U_IR^4. Let p4 = [0]. We unfold the first rule of L4 forwards at position p4 with the rule a__a -> a. ==> L5 = [a__f^#(_0,_0) -> a__g^#(a,a__a), a__g^#(a,_1) -> a__f^#(b,_1)] [comp] is in U_IR^5. Let p5 = [1]. We unfold the first rule of L5 forwards at position p5 with the rule a__a -> b. ==> L6 = a__f^#(_0,_0) -> a__f^#(b,b) [trans] is in U_IR^6. 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 = 1477