TRS Standard pair #615304293

details

property	value
status	complete
benchmark	jwno5.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	Waldmann_06

run statistics

property	value
solver	NTI_23
configuration	default
runtime (wallclock)	0.40233707428 seconds
cpu usage	0.954695877
max memory	9.5707136E7

stage attributes

key	value
output-size	1998
starexec-result	"MAYBE"

output

/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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MAYBE

** 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,f(_1,a)) -> f^#(a,f(f(_0,a),_1)), f^#(_0,f(_1,a)) -> f^#(f(_0,a),_1)]
  TRS = {f(_0,f(_1,a)) -> f(a,f(f(_0,a),_1))}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Failed!
    ## DP Processor: lexicographic path orders. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Failed!
    ## DP Processor: polynomial interpretations. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Failed!
    ## DP Processor: Knuth-Bendix orders. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Failed!
  All the DP processors have 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.
  Failed!
## Could not solve the following DP problems:
1: Dependency pairs = [f^#(_0,f(_1,a)) -> f^#(a,f(f(_0,a),_1)), f^#(_0,f(_1,a)) -> f^#(f(_0,a),_1)]
TRS = {f(_0,f(_1,a)) -> f(a,f(f(_0,a),_1))}
Hence, could not prove (non)termination of the TRS under analysis.
Proof run on Linux version 3.10.0-1160.62.1.el7.x86_64 for amd64
using Java version 1.8.0_322
** END proof description **
Total number of generated unfolded rules = 354

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