/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 = 9] a__f(c,g(c),mark(g(b))) -> a__f(c,g(c),mark(g(b))) 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 = a__f(c,g(c),mark(g(b))) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = a__f(c,g(c),mark(g(b))) 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... ## 2 initial DP problems to solve. ## First, we try to decompose these problems into smaller problems. ## Round 1 [2 DP problems]: ## DP problem: Dependency pairs = [mark^#(g(_0)) -> mark^#(_0)] TRS = {a__f(_0,g(_0),_1) -> a__f(_1,_1,_1), a__g(b) -> c, a__b -> c, mark(f(_0,_1,_2)) -> a__f(_0,_1,_2), mark(g(_0)) -> a__g(mark(_0)), mark(b) -> a__b, mark(c) -> c, a__f(_0,_1,_2) -> f(_0,_1,_2), a__g(_0) -> g(_0), a__b -> b} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [a__f^#(_0,g(_0),_1) -> a__f^#(_1,_1,_1)] TRS = {a__f(_0,g(_0),_1) -> a__f(_1,_1,_1), a__g(b) -> c, a__b -> c, mark(f(_0,_1,_2)) -> a__f(_0,_1,_2), mark(g(_0)) -> a__g(mark(_0)), mark(b) -> a__b, mark(c) -> c, a__f(_0,_1,_2) -> f(_0,_1,_2), a__g(_0) -> g(_0), a__b -> b} ## 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, 1 unfolded rule generated. # Iteration 1: no loop found, 1 unfolded rule generated. # Iteration 2: no loop found, 0 unfolded rule generated. No loop found at all! # max_depth=2, unfold_variables=true: # Iteration 0: no loop found, 1 unfolded rule generated. # Iteration 1: no loop found, 1 unfolded rule generated. # Iteration 2: no loop found, 7 unfolded rules generated. # Iteration 3: no loop found, 11 unfolded rules generated. # Iteration 4: no loop found, 2 unfolded rules generated. # Iteration 5: no loop found, 0 unfolded rule generated. No loop found at all! # max_depth=3, unfold_variables=false: # Iteration 0: no loop found, 1 unfolded rule generated. # Iteration 1: no loop found, 1 unfolded rule generated. # Iteration 2: no loop found, 0 unfolded rule generated. No loop found at all! # max_depth=3, unfold_variables=true: # Iteration 0: no loop found, 1 unfolded rule generated. # Iteration 1: no loop found, 1 unfolded rule generated. # Iteration 2: no loop found, 10 unfolded rules generated. # Iteration 3: no loop found, 20 unfolded rules generated. # Iteration 4: no loop found, 14 unfolded rules generated. # Iteration 5: no loop found, 20 unfolded rules generated. # Iteration 6: no loop found, 22 unfolded rules generated. # Iteration 7: no loop found, 27 unfolded rules generated. # Iteration 8: no loop found, 32 unfolded rules generated. # Iteration 9: success, found a loop, 20 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = a__f^#(_0,g(_0),_1) -> a__f^#(_1,_1,_1) [trans] is in U_IR^0. We build a unit triple from L0. ==> L1 = a__f^#(_0,g(_0),_1) -> a__f^#(_1,_1,_1) [unit] is in U_IR^1. Let p1 = [0]. We unfold the rule of L1 forwards at position p1 with the rule mark(g(_0)) -> a__g(mark(_0)). ==> L2 = a__f^#(_0,g(_0),mark(g(_1))) -> a__f^#(a__g(mark(_1)),mark(g(_1)),mark(g(_1))) [unit] is in U_IR^2. Let p2 = [0, 0]. We unfold the rule of L2 forwards at position p2 with the rule mark(b) -> a__b. ==> L3 = a__f^#(_0,g(_0),mark(g(b))) -> a__f^#(a__g(a__b),mark(g(b)),mark(g(b))) [unit] is in U_IR^3. Let p3 = [0, 0]. We unfold the rule of L3 forwards at position p3 with the rule a__b -> b. ==> L4 = a__f^#(_0,g(_0),mark(g(b))) -> a__f^#(a__g(b),mark(g(b)),mark(g(b))) [unit] is in U_IR^4. Let p4 = [0]. We unfold the rule of L4 forwards at position p4 with the rule a__g(b) -> c. ==> L5 = a__f^#(c,g(c),mark(g(b))) -> a__f^#(c,mark(g(b)),mark(g(b))) [unit] is in U_IR^5. Let p5 = [1]. We unfold the rule of L5 forwards at position p5 with the rule mark(g(_0)) -> a__g(mark(_0)). ==> L6 = a__f^#(c,g(c),mark(g(b))) -> a__f^#(c,a__g(mark(b)),mark(g(b))) [unit] is in U_IR^6. Let p6 = [1]. We unfold the rule of L6 forwards at position p6 with the rule a__g(_0) -> g(_0). ==> L7 = a__f^#(c,g(c),mark(g(b))) -> a__f^#(c,g(mark(b)),mark(g(b))) [unit] is in U_IR^7. Let p7 = [1, 0]. We unfold the rule of L7 forwards at position p7 with the rule mark(b) -> a__b. ==> L8 = a__f^#(c,g(c),mark(g(b))) -> a__f^#(c,g(a__b),mark(g(b))) [unit] is in U_IR^8. Let p8 = [1, 0]. We unfold the rule of L8 forwards at position p8 with the rule a__b -> c. ==> L9 = a__f^#(c,g(c),mark(g(b))) -> a__f^#(c,g(c),mark(g(b))) [unit] is in U_IR^9. 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 = 240