TRS Standard pair #615298693

details

property	value
status	complete
benchmark	test76.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n055.star.cs.uiowa.edu
space	Strategy_removed_mixed_05

run statistics

property	value
solver	NTI_23
configuration	default
runtime (wallclock)	0.796794891357 seconds
cpu usage	2.407612276
max memory	4.49605632E8

stage attributes

key	value
output-size	4657
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: f(g(0,s(0)),s(0),g(0,s(0))) is non-terminating
* Description:
The following rule R was generated while unfolding
the dependency pairs of the analyzed TRS:
[iteration = 5] f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),_0,g(0,_0))
Let l be the left-hand side and r be the right-hand side of R.
Let p = epsilon, theta1 = {_0->s(0)} and theta2 = {}.
We have r|p = f^#(g(0,_0),_0,g(0,_0))
and theta2(theta1(l)) = theta1(r|p), i.e., l semi-unifies with r|p.
So, the term theta1(l) = f(g(0,s(0)),s(0),g(0,s(0)))
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 = [f^#(0,s(0),_0) -> f^#(_0,+(_0,_0),_0)]
  TRS = {+(_0,0) -> _0, +(_0,s(_1)) -> s(+(_0,_1)), f(0,s(0),_0) -> f(_0,+(_0,_0),_0), g(_0,_1) -> _0, g(_0,_1) -> _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 (19 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 = [+^#(_0,s(_1)) -> +^#(_0,_1)]
  TRS = {+(_0,0) -> _0, +(_0,s(_1)) -> s(+(_0,_1)), f(0,s(0),_0) -> f(_0,+(_0,_0),_0), g(_0,_1) -> _0, g(_0,_1) -> _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: infiniteness not proved, 1 unfolded rule generated.
        # Iteration 1: infiniteness not proved, 1 unfolded rule generated.
        # Iteration 2: infiniteness not proved, 7 unfolded rules generated.
        # Iteration 3: infiniteness not proved, 65 unfolded rules generated.
        # Iteration 4: infiniteness not proved, 805 unfolded rules generated.
        # Iteration 5: success, loop found, 6983 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = f^#(0,s(0),_0) -> f^#(_0,+(_0,_0),_0) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = f^#(0,s(0),_0) -> f^#(_0,+(_0,_0),_0) [unit] is in U_IR^1.
        Let p1 = [0].
        We unfold the rule of L1 backwards at position p1
        with the rule g(_0,_1) -> _0.
        ==> L2 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),+(g(0,_0),g(0,_0)),g(0,_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,_1) -> _1.
        ==> L3 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),+(_0,g(0,_0)),g(0,_0)) [unit] is in U_IR^3.
        Let p3 = [1, 1].
        We unfold the rule of L3 forwards at position p3
        with the rule g(_0,_1) -> _0.
        ==> L4 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),+(_0,0),g(0,_0)) [unit] is in U_IR^4.
        Let p4 = [1].
        We unfold the rule of L4 forwards at position p4
        with the rule +(_0,0) -> _0.
        ==> L5 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),_0,g(0,_0)) [unit] is in U_IR^5.
  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 = 7915

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