/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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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 = [f^#(s(_0),_1,_2,_3,_4) -> f^#(_0,_1,_2,_3,_4), f^#(0,s(_0),_1,_2,_3) -> f^#(_0,_0,_1,_2,_3), f^#(0,0,s(_0),_1,_2) -> f^#(_0,_0,_0,_1,_2), f^#(0,0,0,s(_0),_1) -> f^#(_0,_0,_0,_0,_1), f^#(0,0,0,0,s(_0)) -> f^#(_0,_0,_0,_0,_0)]
  TRS = {f(s(_0),_1,_2,_3,_4) -> f(_0,_1,_2,_3,_4), f(0,s(_0),_1,_2,_3) -> f(_0,_0,_1,_2,_3), f(0,0,s(_0),_1,_2) -> f(_0,_0,_0,_1,_2), f(0,0,0,s(_0),_1) -> f(_0,_0,_0,_0,_1), f(0,0,0,0,s(_0)) -> f(_0,_0,_0,_0,_0), f(0,0,0,0,0) -> 0}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... This DP problem is too complex! Aborting!
    ## Trying with lexicographic path orders...
    Successfully decomposed the DP problem into 1 smaller problem to solve!
## Round 2 [1 DP problem]:
  ## DP problem:
  Dependency pairs = [f^#(s(_0),_1,_2,_3,_4) -> f^#(_0,_1,_2,_3,_4), f^#(0,s(_0),_1,_2,_3) -> f^#(_0,_0,_1,_2,_3), f^#(0,0,s(_0),_1,_2) -> f^#(_0,_0,_0,_1,_2), f^#(0,0,0,s(_0),_1) -> f^#(_0,_0,_0,_0,_1)]
  TRS = {f(s(_0),_1,_2,_3,_4) -> f(_0,_1,_2,_3,_4), f(0,s(_0),_1,_2,_3) -> f(_0,_0,_1,_2,_3), f(0,0,s(_0),_1,_2) -> f(_0,_0,_0,_1,_2), f(0,0,0,s(_0),_1) -> f(_0,_0,_0,_0,_1), f(0,0,0,0,s(_0)) -> f(_0,_0,_0,_0,_0), f(0,0,0,0,0) -> 0}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... This DP problem is too complex! Aborting!
    ## Trying with lexicographic path orders...
    Successfully decomposed the DP problem into 1 smaller problem to solve!
## Round 3 [1 DP problem]:
  ## DP problem:
  Dependency pairs = [f^#(s(_0),_1,_2,_3,_4) -> f^#(_0,_1,_2,_3,_4), f^#(0,s(_0),_1,_2,_3) -> f^#(_0,_0,_1,_2,_3), f^#(0,0,s(_0),_1,_2) -> f^#(_0,_0,_0,_1,_2)]
  TRS = {f(s(_0),_1,_2,_3,_4) -> f(_0,_1,_2,_3,_4), f(0,s(_0),_1,_2,_3) -> f(_0,_0,_1,_2,_3), f(0,0,s(_0),_1,_2) -> f(_0,_0,_0,_1,_2), f(0,0,0,s(_0),_1) -> f(_0,_0,_0,_0,_1), f(0,0,0,0,s(_0)) -> f(_0,_0,_0,_0,_0), f(0,0,0,0,0) -> 0}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... This DP problem is too complex! Aborting!
    ## Trying with lexicographic path orders...
    Successfully decomposed the DP problem into 1 smaller problem to solve!
## Round 4 [1 DP problem]:
  ## DP problem:
  Dependency pairs = [f^#(s(_0),_1,_2,_3,_4) -> f^#(_0,_1,_2,_3,_4), f^#(0,s(_0),_1,_2,_3) -> f^#(_0,_0,_1,_2,_3)]
  TRS = {f(s(_0),_1,_2,_3,_4) -> f(_0,_1,_2,_3,_4), f(0,s(_0),_1,_2,_3) -> f(_0,_0,_1,_2,_3), f(0,0,s(_0),_1,_2) -> f(_0,_0,_0,_1,_2), f(0,0,0,s(_0),_1) -> f(_0,_0,_0,_0,_1), f(0,0,0,0,s(_0)) -> f(_0,_0,_0,_0,_0), f(0,0,0,0,0) -> 0}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... This DP problem is too complex! Aborting!
    ## Trying with lexicographic path orders...
    Successfully decomposed the DP problem into 1 smaller problem to solve!
## Round 5 [1 DP problem]:
  ## DP problem:
  Dependency pairs = [f^#(s(_0),_1,_2,_3,_4) -> f^#(_0,_1,_2,_3,_4)]
  TRS = {f(s(_0),_1,_2,_3,_4) -> f(_0,_1,_2,_3,_4), f(0,s(_0),_1,_2,_3) -> f(_0,_0,_1,_2,_3), f(0,0,s(_0),_1,_2) -> f(_0,_0,_0,_1,_2), f(0,0,0,s(_0),_1) -> f(_0,_0,_0,_0,_1), f(0,0,0,0,s(_0)) -> f(_0,_0,_0,_0,_0), f(0,0,0,0,0) -> 0}
    ## Trying with homeomorphic embeddings... Success!
  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