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


--------------------------------------------------------------------------------


YES

** BEGIN proof argument **
All the DP problems were proved finite.
As all the involved DP processors are sound,
the TRS under analysis terminates.
** 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 = [*^#(_0,*(_1,_2)) -> *^#(*(_0,_1),_2), *^#(_0,*(_1,_2)) -> *^#(_0,_1), i^#(*(_0,_1)) -> *^#(i(_1),i(_0)), i^#(*(_0,_1)) -> i^#(_1), i^#(*(_0,_1)) -> i^#(_0), *^#(*(i(_0),k(_1,_2)),_0) -> i^#(_0), *^#(*(i(_0),k(_1,_2)),_0) -> i^#(_0), *^#(*(i(_0),k(_1,_2)),_0) -> *^#(*(i(_0),_1),_0), *^#(*(i(_0),k(_1,_2)),_0) -> *^#(i(_0),_1), *^#(*(i(_0),k(_1,_2)),_0) -> *^#(*(i(_0),_2),_0), *^#(*(i(_0),k(_1,_2)),_0) -> *^#(i(_0),_2)]
  TRS = {*(_0,1) -> _0, *(1,_0) -> _0, *(i(_0),_0) -> 1, *(_0,i(_0)) -> 1, *(_0,*(_1,_2)) -> *(*(_0,_1),_2), i(1) -> 1, *(*(_0,_1),i(_1)) -> _0, *(*(_0,i(_1)),_1) -> _0, i(i(_0)) -> _0, i(*(_0,_1)) -> *(i(_1),i(_0)), k(_0,1) -> 1, k(_0,_0) -> 1, *(k(_0,_1),k(_1,_0)) -> 1, *(*(i(_0),k(_1,_2)),_0) -> k(*(*(i(_0),_1),_0),*(*(i(_0),_2),_0)), k(*(_0,i(_1)),*(_1,i(_0))) -> 1}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations...
    Successfully decomposed the DP problem into 2 smaller problem(s) to solve!
## Round 2 [2 DP problems]:
  ## DP problem:
  Dependency pairs = [i^#(*(_0,_1)) -> i^#(_0), i^#(*(_0,_1)) -> i^#(_1)]
  TRS = {*(_0,1) -> _0, *(1,_0) -> _0, *(i(_0),_0) -> 1, *(_0,i(_0)) -> 1, *(_0,*(_1,_2)) -> *(*(_0,_1),_2), i(1) -> 1, *(*(_0,_1),i(_1)) -> _0, *(*(_0,i(_1)),_1) -> _0, i(i(_0)) -> _0, i(*(_0,_1)) -> *(i(_1),i(_0)), k(_0,1) -> 1, k(_0,_0) -> 1, *(k(_0,_1),k(_1,_0)) -> 1, *(*(i(_0),k(_1,_2)),_0) -> k(*(*(i(_0),_1),_0),*(*(i(_0),_2),_0)), k(*(_0,i(_1)),*(_1,i(_0))) -> 1}
    ## Trying with homeomorphic embeddings... Success!
  This DP problem is finite.
  ## DP problem:
  Dependency pairs = [*^#(_0,*(_1,_2)) -> *^#(*(_0,_1),_2), *^#(_0,*(_1,_2)) -> *^#(_0,_1)]
  TRS = {*(_0,1) -> _0, *(1,_0) -> _0, *(i(_0),_0) -> 1, *(_0,i(_0)) -> 1, *(_0,*(_1,_2)) -> *(*(_0,_1),_2), i(1) -> 1, *(*(_0,_1),i(_1)) -> _0, *(*(_0,i(_1)),_1) -> _0, i(i(_0)) -> _0, i(*(_0,_1)) -> *(i(_1),i(_0)), k(_0,1) -> 1, k(_0,_0) -> 1, *(k(_0,_1),k(_1,_0)) -> 1, *(*(i(_0),k(_1,_2)),_0) -> k(*(*(i(_0),_1),_0),*(*(i(_0),_2),_0)), k(*(_0,i(_1)),*(_1,i(_0))) -> 1}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations...
    Successfully decomposed the DP problem into 1 smaller problem to solve!
## Round 3 [1 DP problem]:
  ## DP problem:
  Dependency pairs = [*^#(_0,*(_1,_2)) -> *^#(*(_0,_1),_2)]
  TRS = {*(_0,1) -> _0, *(1,_0) -> _0, *(i(_0),_0) -> 1, *(_0,i(_0)) -> 1, *(_0,*(_1,_2)) -> *(*(_0,_1),_2), i(1) -> 1, *(*(_0,_1),i(_1)) -> _0, *(*(_0,i(_1)),_1) -> _0, i(i(_0)) -> _0, i(*(_0,_1)) -> *(i(_1),i(_0)), k(_0,1) -> 1, k(_0,_0) -> 1, *(k(_0,_1),k(_1,_0)) -> 1, *(*(i(_0),k(_1,_2)),_0) -> k(*(*(i(_0),_1),_0),*(*(i(_0),_2),_0)), k(*(_0,i(_1)),*(_1,i(_0))) -> 1}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations...
      The constraints are satisfied by the polynomials:
      {i(_0):[_0], k(_0,_1):[_0 + _1], *(_0,_1):[_0 * _1], 1:[2], *^#(_0,_1):[_0]}
      for all instantiations of the variables with values greater than or equal to mu = 2.
  This DP problem is finite.
## All the DP problems were proved finite.
As all the involved DP processors are sound,
the TRS under analysis terminates.
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 = 0