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


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NO

** BEGIN proof argument **
The following rule was generated while unfolding the analyzed TRS:
[iteration = 0] from(_0) -> from(s(_0))
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {} and theta2 = {_0->s(_0)}.
We have r|p = from(s(_0)) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = from(_0) loops 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...
## 4 initial DP problems to solve.
## First, we try to decompose these problems into smaller problems.
## Round 1 [4 DP problems]:
  ## DP problem:
  Dependency pairs = [times^#(s(_0),_1) -> times^#(_0,_1)]
  TRS = {from(_0) -> cons(_0,from(s(_0))), 2ndspos(0,_0) -> rnil, 2ndspos(s(_0),cons(_1,cons(_2,_3))) -> rcons(posrecip(_2),2ndsneg(_0,_3)), 2ndsneg(0,_0) -> rnil, 2ndsneg(s(_0),cons(_1,cons(_2,_3))) -> rcons(negrecip(_2),2ndspos(_0,_3)), pi(_0) -> 2ndspos(_0,from(0)), plus(0,_0) -> _0, plus(s(_0),_1) -> s(plus(_0,_1)), times(0,_0) -> 0, times(s(_0),_1) -> plus(_1,times(_0,_1)), square(_0) -> times(_0,_0)}
    ## Trying with homeomorphic embeddings... Success!
  This DP problem is finite.
  ## DP problem:
  Dependency pairs = [plus^#(s(_0),_1) -> plus^#(_0,_1)]
  TRS = {from(_0) -> cons(_0,from(s(_0))), 2ndspos(0,_0) -> rnil, 2ndspos(s(_0),cons(_1,cons(_2,_3))) -> rcons(posrecip(_2),2ndsneg(_0,_3)), 2ndsneg(0,_0) -> rnil, 2ndsneg(s(_0),cons(_1,cons(_2,_3))) -> rcons(negrecip(_2),2ndspos(_0,_3)), pi(_0) -> 2ndspos(_0,from(0)), plus(0,_0) -> _0, plus(s(_0),_1) -> s(plus(_0,_1)), times(0,_0) -> 0, times(s(_0),_1) -> plus(_1,times(_0,_1)), square(_0) -> times(_0,_0)}
    ## Trying with homeomorphic embeddings... Success!
  This DP problem is finite.
  ## DP problem:
  Dependency pairs = [2ndspos^#(s(_0),cons(_1,cons(_2,_3))) -> 2ndsneg^#(_0,_3), 2ndsneg^#(s(_0),cons(_1,cons(_2,_3))) -> 2ndspos^#(_0,_3)]
  TRS = {from(_0) -> cons(_0,from(s(_0))), 2ndspos(0,_0) -> rnil, 2ndspos(s(_0),cons(_1,cons(_2,_3))) -> rcons(posrecip(_2),2ndsneg(_0,_3)), 2ndsneg(0,_0) -> rnil, 2ndsneg(s(_0),cons(_1,cons(_2,_3))) -> rcons(negrecip(_2),2ndspos(_0,_3)), pi(_0) -> 2ndspos(_0,from(0)), plus(0,_0) -> _0, plus(s(_0),_1) -> s(plus(_0,_1)), times(0,_0) -> 0, times(s(_0),_1) -> plus(_1,times(_0,_1)), square(_0) -> times(_0,_0)}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... This DP problem is too complex! Aborting!
    ## Trying with lexicographic path orders... Too many argument filtering possibilities (46656)! Aborting!
    ## Trying with Knuth-Bendix orders... Failed!
  Don't know whether this DP problem is finite.
  ## DP problem:
  Dependency pairs = [from^#(_0) -> from^#(s(_0))]
  TRS = {from(_0) -> cons(_0,from(s(_0))), 2ndspos(0,_0) -> rnil, 2ndspos(s(_0),cons(_1,cons(_2,_3))) -> rcons(posrecip(_2),2ndsneg(_0,_3)), 2ndsneg(0,_0) -> rnil, 2ndsneg(s(_0),cons(_1,cons(_2,_3))) -> rcons(negrecip(_2),2ndspos(_0,_3)), pi(_0) -> 2ndspos(_0,from(0)), plus(0,_0) -> _0, plus(s(_0),_1) -> s(plus(_0,_1)), times(0,_0) -> 0, times(s(_0),_1) -> plus(_1,times(_0,_1)), square(_0) -> times(_0,_0)}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... This DP problem is too complex! Aborting!
    ## Trying with lexicographic path orders... Too many argument filtering possibilities (46656)! Aborting!
    ## Trying with Knuth-Bendix orders... Failed!
  Don't know whether this DP problem is finite.
## Some DP problems could not be proved finite.
## Now, we try to prove that one of these problems is infinite.
    ## Trying to find a loop (forward=true, backward=true, max=20)
      # max_depth=20, unfold_variables=false:
        # Iteration 0: success, found a loop, 1 unfolded rule generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = from^#(_0) -> from^#(s(_0)) [trans] is in U_IR^0.
  This DP problem is infinite.
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 = 15