/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 = 4] oddNs -> oddNs
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {} and theta2 = {}.
We have r|p = oddNs and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = oddNs 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 = [oddNs^# -> incr^#(pairNs), activate^#(n__oddNs) -> oddNs^#, incr^#(cons(_0,_1)) -> activate^#(_1), activate^#(n__incr(_0)) -> incr^#(activate(_0)), take^#(s(_0),cons(_1,_2)) -> activate^#(_2), activate^#(n__take(_0,_1)) -> take^#(activate(_0),activate(_1)), zip^#(cons(_0,_1),cons(_2,_3)) -> activate^#(_1), activate^#(n__zip(_0,_1)) -> zip^#(activate(_0),activate(_1)), repItems^#(cons(_0,_1)) -> activate^#(_1), activate^#(n__repItems(_0)) -> repItems^#(activate(_0)), activate^#(n__incr(_0)) -> activate^#(_0), activate^#(n__take(_0,_1)) -> activate^#(_0), activate^#(n__take(_0,_1)) -> activate^#(_1), activate^#(n__zip(_0,_1)) -> activate^#(_0), activate^#(n__zip(_0,_1)) -> activate^#(_1), activate^#(n__cons(_0,_1)) -> activate^#(_0), activate^#(n__repItems(_0)) -> activate^#(_0), zip^#(cons(_0,_1),cons(_2,_3)) -> activate^#(_3)]
  TRS = {pairNs -> cons(0,n__incr(n__oddNs)), oddNs -> incr(pairNs), incr(cons(_0,_1)) -> cons(s(_0),n__incr(activate(_1))), take(0,_0) -> nil, take(s(_0),cons(_1,_2)) -> cons(_1,n__take(_0,activate(_2))), zip(nil,_0) -> nil, zip(_0,nil) -> nil, zip(cons(_0,_1),cons(_2,_3)) -> cons(pair(_0,_2),n__zip(activate(_1),activate(_3))), tail(cons(_0,_1)) -> activate(_1), repItems(nil) -> nil, repItems(cons(_0,_1)) -> cons(_0,n__cons(_0,n__repItems(activate(_1)))), incr(_0) -> n__incr(_0), oddNs -> n__oddNs, take(_0,_1) -> n__take(_0,_1), zip(_0,_1) -> n__zip(_0,_1), cons(_0,_1) -> n__cons(_0,_1), repItems(_0) -> n__repItems(_0), activate(n__incr(_0)) -> incr(activate(_0)), activate(n__oddNs) -> oddNs, activate(n__take(_0,_1)) -> take(activate(_0),activate(_1)), activate(n__zip(_0,_1)) -> zip(activate(_0),activate(_1)), activate(n__cons(_0,_1)) -> cons(activate(_0),_1), activate(n__repItems(_0)) -> repItems(activate(_0)), activate(_0) -> _0}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... This DP problem is too complex! Aborting!
    ## Trying with lexicographic path orders... Too many argument filtering possibilities (279936)! Aborting!
    ## Trying with Knuth-Bendix orders... This DP problem is too complex! Aborting!
  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, 18 unfolded rules generated.
        # Iteration 1: no loop found, 282 unfolded rules generated.
        # Iteration 2: no loop found, 1746 unfolded rules generated.
        # Iteration 3: no loop found, 5986 unfolded rules generated.
        # Iteration 4: success, found a loop, 8 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = oddNs^# -> incr^#(pairNs) [trans] is in U_IR^0.
        D = incr^#(cons(_0,_1)) -> activate^#(_1) is a dependency pair of IR.
        We build a composed triple from L0 and D.
        ==> L1 = [oddNs^# -> incr^#(pairNs), incr^#(cons(_0,_1)) -> activate^#(_1)] [comp] is in U_IR^1.
        Let p1 = [0].
        We unfold the first rule of L1 forwards at position p1
        with the rule pairNs -> cons(0,n__incr(n__oddNs)).
        ==> L2 = oddNs^# -> activate^#(n__incr(n__oddNs)) [trans] is in U_IR^2.
        D = activate^#(n__incr(_0)) -> activate^#(_0) is a dependency pair of IR.
        We build a composed triple from L2 and D.
        ==> L3 = oddNs^# -> activate^#(n__oddNs) [trans] is in U_IR^3.
        D = activate^#(n__oddNs) -> oddNs^# is a dependency pair of IR.
        We build a composed triple from L3 and D.
        ==> L4 = oddNs^# -> oddNs^# [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 = 19636