Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Standard pair #516966792
details
property
value
status
complete
benchmark
4.34.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n029.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
NTI_22
configuration
default
runtime (wallclock)
0.2341401577 seconds
cpu usage
0.253011406
max memory
3.9387136E7
stage attributes
key
value
output-size
2623
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 = 2] a(b(b(_0))) -> a(b(b(a(a(_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->a(a(_0))}. We have r|p = a(b(b(a(a(_0))))) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = a(b(b(_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^#(b(_0)) -> a^#(a(_0)), a^#(b(_0)) -> a^#(_0)] TRS = {a(b(_0)) -> b(b(a(a(_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. ## 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=4, unfold_variables=false: # Iteration 0: no loop found, 2 unfolded rules generated. # Iteration 1: no loop found, 4 unfolded rules generated. # Iteration 2: success, found a loop, 1 unfolded rule generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = a^#(b(_0)) -> a^#(a(_0)) [trans] is in U_IR^0. D = a^#(b(_0)) -> a^#(_0) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = [a^#(b(_0)) -> a^#(a(_0)), a^#(b(_1)) -> a^#(_1)] [comp] is in U_IR^1. Let p1 = [0]. We unfold the first rule of L1 forwards at position p1 with the rule a(b(_0)) -> b(b(a(a(_0)))). ==> L2 = [a^#(b(b(_0))) -> a^#(b(b(a(a(_0))))), a^#(b(_1)) -> a^#(_1)] [comp] is in U_IR^2. 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 = 22
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Standard