/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] activate(n__f(n__g(g(_0)),_1)) -> activate(n__f(n__g(g(_0)),activate(_1)))
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {} and theta2 = {_1->activate(_1)}.
We have r|p = activate(n__f(n__g(g(_0)),activate(_1))) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = activate(n__f(n__g(g(_0)),_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^#(g(_0),_1) -> f^#(_0,n__f(n__g(_0),activate(_1))), activate^#(n__f(_0,_1)) -> f^#(activate(_0),_1), activate^#(n__f(_0,_1)) -> activate^#(_0), activate^#(n__g(_0)) -> activate^#(_0), f^#(g(_0),_1) -> activate^#(_1)]
  TRS = {f(g(_0),_1) -> f(_0,n__f(n__g(_0),activate(_1))), f(_0,_1) -> n__f(_0,_1), g(_0) -> n__g(_0), activate(n__f(_0,_1)) -> f(activate(_0),_1), activate(n__g(_0)) -> g(activate(_0)), activate(_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.
## 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=3, unfold_variables=false:
        # Iteration 0: no loop found, 5 unfolded rules generated.
        # Iteration 1: no loop found, 18 unfolded rules generated.
        # Iteration 2: no loop found, 31 unfolded rules generated.
        # Iteration 3: no loop found, 34 unfolded rules generated.
        # Iteration 4: success, found a loop, 10 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = activate^#(n__f(_0,_1)) -> f^#(activate(_0),_1) [trans] is in U_IR^0.
        D = f^#(g(_0),_1) -> f^#(_0,n__f(n__g(_0),activate(_1))) is a dependency pair of IR.
        We build a composed triple from L0 and D.
        ==> L1 = [activate^#(n__f(_0,_1)) -> f^#(activate(_0),_1), f^#(g(_2),_3) -> f^#(_2,n__f(n__g(_2),activate(_3)))] [comp] is in U_IR^1.
        Let p1 = [0].
        We unfold the first rule of L1 forwards at position p1
        with the rule activate(n__g(_0)) -> g(activate(_0)).
        ==> L2 = [activate^#(n__f(n__g(_0),_1)) -> f^#(g(activate(_0)),_1), f^#(g(_2),_3) -> f^#(_2,n__f(n__g(_2),activate(_3)))] [comp] is in U_IR^2.
        Let p2 = [0, 0].
        We unfold the first rule of L2 forwards at position p2
        with the rule activate(_0) -> _0.
        ==> L3 = activate^#(n__f(n__g(_0),_1)) -> f^#(_0,n__f(n__g(_0),activate(_1))) [trans] is in U_IR^3.
        D = f^#(g(_0),_1) -> activate^#(_1) is a dependency pair of IR.
        We build a composed triple from L3 and D.
        ==> L4 = activate^#(n__f(n__g(g(_0)),_1)) -> activate^#(n__f(n__g(g(_0)),activate(_1))) [trans] 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 = 438