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TRS Standard pair #516967697
details
property
value
status
complete
benchmark
#4.14.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n068.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
NTI_22
configuration
default
runtime (wallclock)
0.333407878876 seconds
cpu usage
0.445332922
max memory
4.5658112E7
stage attributes
key
value
output-size
3559
starexec-result
NO
output
/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 = 3] f(g(s(0)),s(0)) -> f(g(s(0)),s(0)) 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(g(s(0)),s(0)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = f(g(s(0)),s(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... ## 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 = [f^#(g(_0),s(0)) -> f^#(g(_0),g(_0))] TRS = {f(g(_0),s(0)) -> f(g(_0),g(_0)), g(s(_0)) -> s(g(_0)), g(0) -> 0} ## 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. ## DP problem: Dependency pairs = [g^#(s(_0)) -> g^#(_0)] TRS = {f(g(_0),s(0)) -> f(g(_0),g(_0)), g(s(_0)) -> s(g(_0)), g(0) -> 0} ## Trying with homeomorphic embeddings... Success! 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, 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, 1 unfolded rule generated. # Iteration 3: success, found a loop, 1 unfolded rule generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = f^#(g(_0),s(0)) -> f^#(g(_0),g(_0)) [trans] is in U_IR^0. We build a unit triple from L0. ==> L1 = f^#(g(_0),s(0)) -> f^#(g(_0),g(_0)) [unit] is in U_IR^1. Let p1 = [1]. We unfold the rule of L1 forwards at position p1 with the rule g(s(_0)) -> s(g(_0)). ==> L2 = f^#(g(s(_0)),s(0)) -> f^#(g(s(_0)),s(g(_0))) [unit] is in U_IR^2. Let p2 = [1, 0]. We unfold the rule of L2 forwards at position p2 with the rule g(0) -> 0. ==> L3 = f^#(g(s(0)),s(0)) -> f^#(g(s(0)),s(0)) [unit] is in U_IR^3. 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 = 29
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