/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 = 3] f(_0,h(h(_1))) -> f(h(h(f(f(h(a),a),f(h(a),_1)))),_0)
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {_0->h(h(_2))} and theta2 = {_1->_2, _2->f(f(h(a),a),f(h(a),_1))}.
We have r|p = f(h(h(f(f(h(a),a),f(h(a),_1)))),_0) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = f(h(h(_2)),h(h(_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,h(_1)) -> f^#(f(h(a),_1),_0), f^#(_0,h(_1)) -> f^#(h(a),_1)]
  TRS = {f(_0,h(_1)) -> h(f(f(h(a),_1),_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=true, max=20)
      # max_depth=20, unfold_variables=false:
        # Iteration 0: no loop found, 2 unfolded rules generated.
        # Iteration 1: no loop found, 5 unfolded rules generated.
        # Iteration 2: no loop found, 8 unfolded rules generated.
        # Iteration 3: success, found a loop, 1 unfolded rule generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = f^#(_0,h(_1)) -> f^#(f(h(a),_1),_0) [trans] is in U_IR^0.
        D = f^#(_0,h(_1)) -> f^#(h(a),_1) is a dependency pair of IR.
        We build a composed triple from L0 and D.
        ==> L1 = [f^#(_0,h(_1)) -> f^#(f(h(a),_1),_0), f^#(_2,h(_3)) -> f^#(h(a),_3)] [comp] is in U_IR^1.
        Let p1 = [0].
        We unfold the first rule of L1 forwards at position p1
        with the rule f(_0,h(_1)) -> h(f(f(h(a),_1),_0)).
        ==> L2 = [f^#(_0,h(h(_1))) -> f^#(h(f(f(h(a),_1),h(a))),_0), f^#(_2,h(_3)) -> f^#(h(a),_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 f(_0,h(_1)) -> h(f(f(h(a),_1),_0)).
        ==> L3 = [f^#(_0,h(h(_1))) -> f^#(h(h(f(f(h(a),a),f(h(a),_1)))),_0), f^#(_2,h(_3)) -> f^#(h(a),_3)] [comp] is in U_IR^3.
  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 = 61