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


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NO

** BEGIN proof argument **
The following rule was generated while unfolding the analyzed TRS:
[iteration = 5] f(g(0,_0,_0),g(2,1,1),g(0,_0,_0)) -> f(g(0,_0,_0),g(2,_0,_0),g(0,_0,_0))
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {_0->1} and theta2 = {}.
We have r|p = f(g(0,_0,_0),g(2,_0,_0),g(0,_0,_0)) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = f(g(0,1,1),g(2,1,1),g(0,1,1)) 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...
## 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^#(0,1,_0) -> f^#(_0,_0,_0)]
  TRS = {f(0,1,_0) -> f(_0,_0,_0), f(_0,_1,_2) -> 2, 0 -> 2, 1 -> 2, g(_0,_0,_1) -> _1, g(_0,_1,_1) -> _0}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... This DP problem is too complex! Aborting!
    ## Trying with lexicographic path orders... Failed!
    ## Trying with Knuth-Bendix orders... 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.
    ## Trying to find a loop (forward=true, backward=true, max=20)
      # max_depth=20, unfold_variables=false:
        # Iteration 0: no loop found, 1 unfolded rule generated.
        # Iteration 1: no loop found, 1 unfolded rule generated.
        # Iteration 2: no loop found, 5 unfolded rules generated.
        # Iteration 3: no loop found, 24 unfolded rules generated.
        # Iteration 4: no loop found, 138 unfolded rules generated.
        # Iteration 5: success, found a loop, 1044 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = f^#(0,1,_0) -> f^#(_0,_0,_0) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = f^#(0,1,_0) -> f^#(_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,_1) -> _0.
        ==> L2 = f^#(g(0,_0,_0),1,g(0,_0,_0)) -> f^#(g(0,_0,_0),g(0,_0,_0),g(0,_0,_0)) [unit] is in U_IR^2.
        Let p2 = [1].
        We unfold the rule of L2 backwards at position p2
        with the rule g(_0,_0,_1) -> _1.
        ==> L3 = f^#(g(0,_0,_0),g(_1,_1,1),g(0,_0,_0)) -> f^#(g(0,_0,_0),g(0,_0,_0),g(0,_0,_0)) [unit] is in U_IR^3.
        Let p3 = [1, 0].
        We unfold the rule of L3 forwards at position p3
        with the rule 0 -> 2.
        ==> L4 = f^#(g(0,_0,_0),g(2,2,1),g(0,_0,_0)) -> f^#(g(0,_0,_0),g(2,_0,_0),g(0,_0,_0)) [unit] is in U_IR^4.
        Let p4 = [1, 1].
        We unfold the rule of L4 backwards at position p4
        with the rule 1 -> 2.
        ==> L5 = f^#(g(0,_0,_0),g(2,1,1),g(0,_0,_0)) -> f^#(g(0,_0,_0),g(2,_0,_0),g(0,_0,_0)) [unit] is in U_IR^5.
  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 = 1223