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TRS Standard pair #516962567
details
property
value
status
complete
benchmark
Ex24_GM04_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
NTI_22
configuration
default
runtime (wallclock)
0.323997974396 seconds
cpu usage
0.56794201
max memory
6.0522496E7
stage attributes
key
value
output-size
5063
starexec-result
NO
output
/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] f(c,n__g(c),n__g(b)) -> f(c,n__g(c),n__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 = f(c,n__g(c),n__g(b)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = f(c,n__g(c),n__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... ## 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,n__g(_0),_1) -> f^#(activate(_1),activate(_1),activate(_1))] TRS = {f(_0,n__g(_0),_1) -> f(activate(_1),activate(_1),activate(_1)), g(b) -> c, b -> c, g(_0) -> n__g(_0), activate(n__g(_0)) -> g(_0), activate(_0) -> _0} ## 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, 2 unfolded rules generated. # Iteration 3: no loop found, 2 unfolded rules generated. # Iteration 4: no loop found, 3 unfolded rules generated. # Iteration 5: no loop found, 2 unfolded rules generated. # Iteration 6: 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, 2 unfolded rules generated. # Iteration 3: no loop found, 2 unfolded rules generated. # Iteration 4: no loop found, 1 unfolded rule generated. # Iteration 5: no loop found, 2 unfolded rules generated. # Iteration 6: no loop found, 6 unfolded rules generated. # Iteration 7: 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, 5 unfolded rules generated. # Iteration 3: no loop found, 13 unfolded rules generated. # Iteration 4: no loop found, 20 unfolded rules generated. # Iteration 5: no loop found, 22 unfolded rules generated. # Iteration 6: success, found a loop, 14 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = f^#(_0,n__g(_0),_1) -> f^#(activate(_1),activate(_1),activate(_1)) [trans] is in U_IR^0. We build a unit triple from L0. ==> L1 = f^#(_0,n__g(_0),_1) -> f^#(activate(_1),activate(_1),activate(_1)) [unit] is in U_IR^1. Let p1 = [0]. We unfold the rule of L1 forwards at position p1 with the rule activate(n__g(_0)) -> g(_0). ==> L2 = f^#(_0,n__g(_0),n__g(_1)) -> f^#(g(_1),activate(n__g(_1)),activate(n__g(_1))) [unit] is in U_IR^2. Let p2 = [0]. We unfold the rule of L2 forwards at position p2 with the rule g(b) -> c. ==> L3 = f^#(c,n__g(c),n__g(b)) -> f^#(c,activate(n__g(b)),activate(n__g(b))) [unit] is in U_IR^3. Let p3 = [1]. We unfold the rule of L3 forwards at position p3 with the rule activate(_0) -> _0. ==> L4 = f^#(c,n__g(c),n__g(b)) -> f^#(c,n__g(b),activate(n__g(b))) [unit] is in U_IR^4. Let p4 = [1, 0]. We unfold the rule of L4 forwards at position p4 with the rule b -> c. ==> L5 = f^#(c,n__g(c),n__g(b)) -> f^#(c,n__g(c),activate(n__g(b))) [unit] is in U_IR^5. Let p5 = [2]. We unfold the rule of L5 forwards at position p5 with the rule activate(_0) -> _0. ==> L6 = f^#(c,n__g(c),n__g(b)) -> f^#(c,n__g(c),n__g(b)) [unit] 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 = 568
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