/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


NO

** BEGIN proof argument **
The following rule was generated while unfolding the analyzed TRS:
[iteration = 3] length(cons(_0,n__zeros)) -> length(cons(0,n__zeros))
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {} and theta2 = {_0->0}.
We have r|p = length(cons(0,n__zeros)) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = length(cons(_0,n__zeros)) 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^#(cons(_0,_1)) -> length^#(activate(_1))]
  TRS = {zeros -> cons(0,n__zeros), and(tt,_0) -> activate(_0), length(nil) -> 0, length(cons(_0,_1)) -> s(length(activate(_1))), take(0,_0) -> nil, take(s(_0),cons(_1,_2)) -> cons(_1,n__take(_0,activate(_2))), zeros -> n__zeros, take(_0,_1) -> n__take(_0,_1), activate(n__zeros) -> zeros, activate(n__take(_0,_1)) -> take(activate(_0),activate(_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__take(_0,_1)) -> take^#(activate(_0),activate(_1)), activate^#(n__take(_0,_1)) -> activate^#(_0), activate^#(n__take(_0,_1)) -> activate^#(_1), take^#(s(_0),cons(_1,_2)) -> activate^#(_2)]
  TRS = {zeros -> cons(0,n__zeros), and(tt,_0) -> activate(_0), length(nil) -> 0, length(cons(_0,_1)) -> s(length(activate(_1))), take(0,_0) -> nil, take(s(_0),cons(_1,_2)) -> cons(_1,n__take(_0,activate(_2))), zeros -> n__zeros, take(_0,_1) -> n__take(_0,_1), activate(n__zeros) -> zeros, activate(n__take(_0,_1)) -> take(activate(_0),activate(_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.
## Some DP problems could not be proved finite.
## Now, we try to prove that one of these problems is infinite.
    ## Trying to find a loop (forward=true, backward=false, max=-1)
      # 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, 2 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 = length^#(cons(_0,_1)) -> length^#(activate(_1)) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = length^#(cons(_0,_1)) -> length^#(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__zeros) -> zeros.
        ==> L2 = length^#(cons(_0,n__zeros)) -> length^#(zeros) [unit] is in U_IR^2.
        Let p2 = [0].
        We unfold the rule of L2 forwards at position p2
        with the rule zeros -> cons(0,n__zeros).
        ==> L3 = length^#(cons(_0,n__zeros)) -> length^#(cons(0,n__zeros)) [unit] 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 = 81