/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 = 7] f(c,n__g(c),n__g(b)) -> f(c,n__g(c),n__g(b))
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(c,n__g(c),n__g(b)) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = f(c,n__g(c),n__g(b)) 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^#(_0,n__g(_0),_1) -> f^#(activate(_1),activate(_1),activate(_1))]
  TRS = {f(_0,n__g(_0),_1) -> f(activate(_1),activate(_1),activate(_1)), g(b) -> c, b -> c, g(_0) -> n__g(_0), activate(n__g(_0)) -> g(activate(_0)), activate(_0) -> _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.
  ## DP problem:
  Dependency pairs = [activate^#(n__g(_0)) -> activate^#(_0)]
  TRS = {f(_0,n__g(_0),_1) -> f(activate(_1),activate(_1),activate(_1)), g(b) -> c, b -> c, g(_0) -> n__g(_0), activate(n__g(_0)) -> g(activate(_0)), activate(_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=false, max=-1)
      # max_depth=2, 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, 2 unfolded rules generated.
        # Iteration 3: no loop found, 2 unfolded rules generated.
        # Iteration 4: no loop found, 2 unfolded rules generated.
        # Iteration 5: no loop found, 0 unfolded rule generated.
        No loop found at all!
      # max_depth=2, unfold_variables=true:
        # Iteration 0: no loop found, 1 unfolded rule generated.
        # Iteration 1: no loop found, 1 unfolded rule generated.
        # Iteration 2: no loop found, 2 unfolded rules generated.
        # Iteration 3: no loop found, 2 unfolded rules generated.
        # Iteration 4: no loop found, 1 unfolded rule generated.
        # Iteration 5: no loop found, 2 unfolded rules generated.
        # Iteration 6: no loop found, 6 unfolded rules generated.
        # Iteration 7: no loop found, 0 unfolded rule generated.
        No loop found at all!
      # max_depth=3, 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, 4 unfolded rules generated.
        # Iteration 3: no loop found, 8 unfolded rules generated.
        # Iteration 4: no loop found, 16 unfolded rules generated.
        # Iteration 5: no loop found, 19 unfolded rules generated.
        # Iteration 6: no loop found, 31 unfolded rules generated.
        # Iteration 7: success, found a loop, 23 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = f^#(_0,n__g(_0),_1) -> f^#(activate(_1),activate(_1),activate(_1)) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = f^#(_0,n__g(_0),_1) -> f^#(activate(_1),activate(_1),activate(_1)) [unit] is in U_IR^1.
        Let p1 = [0].
        We unfold the rule of L1 forwards at position p1
        with the rule activate(n__g(_0)) -> g(activate(_0)).
        ==> L2 = f^#(_0,n__g(_0),n__g(_1)) -> f^#(g(activate(_1)),activate(n__g(_1)),activate(n__g(_1))) [unit] is in U_IR^2.
        Let p2 = [0, 0].
        We unfold the rule of L2 forwards at position p2
        with the rule activate(_0) -> _0.
        ==> L3 = f^#(_0,n__g(_0),n__g(_1)) -> f^#(g(_1),activate(n__g(_1)),activate(n__g(_1))) [unit] is in U_IR^3.
        Let p3 = [0].
        We unfold the rule of L3 forwards at position p3
        with the rule g(b) -> c.
        ==> L4 = f^#(c,n__g(c),n__g(b)) -> f^#(c,activate(n__g(b)),activate(n__g(b))) [unit] is in U_IR^4.
        Let p4 = [1].
        We unfold the rule of L4 forwards at position p4
        with the rule activate(_0) -> _0.
        ==> L5 = f^#(c,n__g(c),n__g(b)) -> f^#(c,n__g(b),activate(n__g(b))) [unit] is in U_IR^5.
        Let p5 = [1, 0].
        We unfold the rule of L5 forwards at position p5
        with the rule b -> c.
        ==> L6 = f^#(c,n__g(c),n__g(b)) -> f^#(c,n__g(c),activate(n__g(b))) [unit] is in U_IR^6.
        Let p6 = [2].
        We unfold the rule of L6 forwards at position p6
        with the rule activate(_0) -> _0.
        ==> L7 = f^#(c,n__g(c),n__g(b)) -> f^#(c,n__g(c),n__g(b)) [unit] is in U_IR^7.
  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 = 470