/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 = 4] f(g(s(0)),s(0),g(s(0))) -> f(g(s(0)),s(0),g(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 = {}.
We have r|p = f(g(s(0)),s(0),g(s(0))) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = f(g(s(0)),s(0),g(s(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...
## 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^#(g(_0),s(0),_1) -> f^#(_1,_1,g(_0))]
  TRS = {f(g(_0),s(0),_1) -> f(_1,_1,g(_0)), g(s(_0)) -> s(g(_0)), g(0) -> 0}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... Failed!
    ## Trying with lexicographic path orders... Failed!
    ## Trying with Knuth-Bendix orders... Failed!
  Don't know whether this DP problem is finite.
  ## DP problem:
  Dependency pairs = [g^#(s(_0)) -> g^#(_0)]
  TRS = {f(g(_0),s(0),_1) -> f(_1,_1,g(_0)), g(s(_0)) -> s(g(_0)), g(0) -> 0}
    ## Trying with homeomorphic embeddings... Success!
  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, 1 unfolded rule generated.
        # Iteration 3: no loop found, 1 unfolded rule generated.
        # Iteration 4: success, found a loop, 1 unfolded rule generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = f^#(g(_0),s(0),_1) -> f^#(_1,_1,g(_0)) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = f^#(g(_0),s(0),_1) -> f^#(_1,_1,g(_0)) [unit] is in U_IR^1.
        Let p1 = [0].
        The subterm at position p1 in the left-hand side of the rule of L1 unifies with
        the subterm at position p1 in the right-hand side of the rule of L1.
        ==> L2 = f^#(g(_0),s(0),g(_0)) -> f^#(g(_0),g(_0),g(_0)) [unit] is in U_IR^2.
        Let p2 = [1].
        We unfold the rule of L2 forwards at position p2
        with the rule g(s(_0)) -> s(g(_0)).
        ==> L3 = f^#(g(s(_0)),s(0),g(s(_0))) -> f^#(g(s(_0)),s(g(_0)),g(s(_0))) [unit] is in U_IR^3.
        Let p3 = [1, 0].
        We unfold the rule of L3 forwards at position p3
        with the rule g(0) -> 0.
        ==> L4 = f^#(g(s(0)),s(0),g(s(0))) -> f^#(g(s(0)),s(0),g(s(0))) [unit] is in U_IR^4.
  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 = 22