TRS Standard pair #615306855

details

property	value
status	complete
benchmark	Ex1_GM03.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n150.star.cs.uiowa.edu
space	Strategy_removed_CSR_05

run statistics

property	value
solver	NTI_23
configuration	default
runtime (wallclock)	0.469362020493 seconds
cpu usage	0.944873502
max memory	6.1943808E7

stage attributes

key	value
output-size	3399
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 **
* Certificate: diff(_0,_1) is non-terminating
* Description:
The following rule R was generated while unfolding
the dependency pairs of the analyzed TRS:
[iteration = 0] diff^#(_0,_1) -> diff^#(p(_0),_1)
Let l be the left-hand side and r be the right-hand side of R.
Let p = epsilon, theta1 = {} and theta2 = {_0->p(_0)}.
We have r|p = diff^#(p(_0),_1)
and theta2(theta1(l)) = theta1(r|p), i.e., l semi-unifies with r|p.
So, the term theta1(l) = diff(_0,_1)
starts an infinite rewrite sequence 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 = [diff^#(_0,_1) -> diff^#(p(_0),_1)]
  TRS = {p(0) -> 0, p(s(_0)) -> _0, leq(0,_0) -> true, leq(s(_0),0) -> false, leq(s(_0),s(_1)) -> leq(_0,_1), if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, diff(_0,_1) -> if(leq(_0,_1),0,s(diff(p(_0),_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.
      This DP problem is too complex (23 coefficients to instantiate)! Aborting!
      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.
  ## DP problem:
  Dependency pairs = [leq^#(s(_0),s(_1)) -> leq^#(_0,_1)]
  TRS = {p(0) -> 0, p(s(_0)) -> _0, leq(0,_0) -> true, leq(s(_0),0) -> false, leq(s(_0),s(_1)) -> leq(_0,_1), if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, diff(_0,_1) -> if(leq(_0,_1),0,s(diff(p(_0),_1)))}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Success, finiteness proved!
  This DP problem is finite.
## A DP problem could not be proved finite.
## Now, we try to prove that this problem is infinite.
    ## DP Processor: [Payet, LOPSTR'18] + [Payet, Non-termination in TRS and LP] (forward=true, backward=true, max=20). 
      # max_depth=20, unfold_variables=false:
        # Iteration 0: success, loop found, 1 unfolded rule generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = diff^#(_0,_1) -> diff^#(p(_0),_1) [trans] is in U_IR^0.
  This DP problem is infinite.
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 = 3

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