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