TRS Standard pair #615302501

details

property	value
status	complete
benchmark	LengthOfFiniteLists_nosorts_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n164.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	NTI_23
configuration	default
runtime (wallclock)	0.851002931595 seconds
cpu usage	2.622089803
max memory	1.24461056E8

stage attributes

key	value
output-size	4553
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 **
* Certificate: a__length(cons(_0,zeros)) is non-terminating
* Description:
The following rule R was generated while unfolding
the dependency pairs of the analyzed TRS:
[iteration = 3] a__length^#(cons(_0,zeros)) -> a__length^#(cons(0,zeros))
Let l be the left-hand side and r be the right-hand side of R.
Let p = epsilon, theta1 = {} and theta2 = {_0->0}.
We have r|p = a__length^#(cons(0,zeros))
and theta2(theta1(l)) = theta1(r|p), i.e., l semi-unifies with r|p.
So, the term theta1(l) = a__length(cons(_0,zeros))
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...
## 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__and^#(tt,_0) -> mark^#(_0), mark^#(and(_0,_1)) -> a__and^#(mark(_0),_1), a__length^#(cons(_0,_1)) -> mark^#(_1), a__length^#(cons(_0,_1)) -> a__length^#(mark(_1)), mark^#(length(_0)) -> a__length^#(mark(_0)), mark^#(and(_0,_1)) -> mark^#(_0), mark^#(length(_0)) -> mark^#(_0), mark^#(cons(_0,_1)) -> mark^#(_0), mark^#(s(_0)) -> mark^#(_0)]
  TRS = {a__zeros -> cons(0,zeros), a__and(tt,_0) -> mark(_0), a__length(nil) -> 0, a__length(cons(_0,_1)) -> s(a__length(mark(_1))), mark(zeros) -> a__zeros, mark(and(_0,_1)) -> a__and(mark(_0),_1), mark(length(_0)) -> a__length(mark(_0)), mark(cons(_0,_1)) -> cons(mark(_0),_1), mark(0) -> 0, mark(tt) -> tt, mark(nil) -> nil, mark(s(_0)) -> s(mark(_0)), a__zeros -> zeros, a__and(_0,_1) -> and(_0,_1), a__length(_0) -> length(_0)}
    ## 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 (25 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.
## 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=false, max=-1). 
      # max_depth=3, unfold_variables=false:
        # Iteration 0: infiniteness not proved, 9 unfolded rules generated.
        # Iteration 1: infiniteness not proved, 69 unfolded rules generated.
        # Iteration 2: infiniteness not proved, 154 unfolded rules generated.
        # Iteration 3: success, loop found, 96 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = a__length^#(cons(_0,_1)) -> a__length^#(mark(_1)) [trans] is in U_IR^0.
        D = a__length^#(cons(_0,_1)) -> mark^#(_1) is a dependency pair of IR.
        We build a composed triple from L0 and D.
        ==> L1 = [a__length^#(cons(_0,_1)) -> a__length^#(mark(_1)), a__length^#(cons(_2,_3)) -> mark^#(_3)] [comp] is in U_IR^1.
        Let p1 = [0].
        We unfold the first rule of L1 forwards at position p1
        with the rule mark(zeros) -> a__zeros.
        ==> L2 = [a__length^#(cons(_0,zeros)) -> a__length^#(a__zeros), a__length^#(cons(_1,_2)) -> mark^#(_2)] [comp] is in U_IR^2.
        Let p2 = [0].
        We unfold the first rule of L2 forwards at position p2
        with the rule a__zeros -> cons(0,zeros).
        ==> L3 = [a__length^#(cons(_0,zeros)) -> a__length^#(cons(0,zeros)), a__length^#(cons(_1,_2)) -> mark^#(_2)] [comp] is in U_IR^3.
  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 = 653

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