/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 = 6] length(n__cons(_0,n__from(_1))) -> length(n__cons(activate(_1),n__from(n__s(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 = {_0->activate(_1), _1->n__s(activate(_1))}.
We have r|p = length(n__cons(activate(_1),n__from(n__s(activate(_1))))) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = length(n__cons(_0,n__from(_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...
## 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 = [length^#(n__cons(_0,_1)) -> length1^#(activate(_1)), length1^#(_0) -> length^#(activate(_0))]
  TRS = {from(_0) -> cons(_0,n__from(n__s(_0))), length(n__nil) -> 0, length(n__cons(_0,_1)) -> s(length1(activate(_1))), length1(_0) -> length(activate(_0)), from(_0) -> n__from(_0), s(_0) -> n__s(_0), nil -> n__nil, cons(_0,_1) -> n__cons(_0,_1), activate(n__from(_0)) -> from(activate(_0)), activate(n__s(_0)) -> s(activate(_0)), activate(n__nil) -> nil, activate(n__cons(_0,_1)) -> cons(activate(_0),_1), 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__from(_0)) -> activate^#(_0), activate^#(n__s(_0)) -> activate^#(_0), activate^#(n__cons(_0,_1)) -> activate^#(_0)]
  TRS = {from(_0) -> cons(_0,n__from(n__s(_0))), length(n__nil) -> 0, length(n__cons(_0,_1)) -> s(length1(activate(_1))), length1(_0) -> length(activate(_0)), from(_0) -> n__from(_0), s(_0) -> n__s(_0), nil -> n__nil, cons(_0,_1) -> n__cons(_0,_1), activate(n__from(_0)) -> from(activate(_0)), activate(n__s(_0)) -> s(activate(_0)), activate(n__nil) -> nil, activate(n__cons(_0,_1)) -> cons(activate(_0),_1), 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=true, max=20)
      # max_depth=20, unfold_variables=false:
        # Iteration 0: no loop found, 2 unfolded rules generated.
        # Iteration 1: no loop found, 3 unfolded rules generated.
        # Iteration 2: no loop found, 12 unfolded rules generated.
        # Iteration 3: no loop found, 8 unfolded rules generated.
        # Iteration 4: no loop found, 64 unfolded rules generated.
        # Iteration 5: no loop found, 566 unfolded rules generated.
        # Iteration 6: success, found a loop, 1 unfolded rule generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = length^#(n__cons(_0,_1)) -> length1^#(activate(_1)) [trans] is in U_IR^0.
        D = length1^#(_0) -> length^#(activate(_0)) is a dependency pair of IR.
        We build a composed triple from L0 and D.
        ==> L1 = length^#(n__cons(_0,_1)) -> length^#(activate(activate(_1))) [trans] is in U_IR^1.
        We build a unit triple from L1.
        ==> L2 = length^#(n__cons(_0,_1)) -> length^#(activate(activate(_1))) [unit] is in U_IR^2.
        Let p2 = [0].
        We unfold the rule of L2 forwards at position p2
        with the rule activate(_0) -> _0.
        ==> L3 = length^#(n__cons(_0,_1)) -> length^#(activate(_1)) [unit] is in U_IR^3.
        Let p3 = [0].
        We unfold the rule of L3 forwards at position p3
        with the rule activate(n__from(_0)) -> from(activate(_0)).
        ==> L4 = length^#(n__cons(_0,n__from(_1))) -> length^#(from(activate(_1))) [unit] is in U_IR^4.
        Let p4 = [0].
        We unfold the rule of L4 forwards at position p4
        with the rule from(_0) -> cons(_0,n__from(n__s(_0))).
        ==> L5 = length^#(n__cons(_0,n__from(_1))) -> length^#(cons(activate(_1),n__from(n__s(activate(_1))))) [unit] is in U_IR^5.
        Let p5 = [0].
        We unfold the rule of L5 forwards at position p5
        with the rule cons(_0,_1) -> n__cons(_0,_1).
        ==> L6 = length^#(n__cons(_0,n__from(_1))) -> length^#(n__cons(activate(_1),n__from(n__s(activate(_1))))) [unit] is in U_IR^6.
  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 = 997