/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] sieve(cons(_0,n__from(_1))) -> sieve(cons(_1,n__from(s(_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->_1, _1->s(_1)}.
We have r|p = sieve(cons(_1,n__from(s(_1)))) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = sieve(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...
## 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 = [sieve^#(cons(_0,_1)) -> sieve^#(activate(_1)), filter^#(s(s(_0)),cons(_1,_2)) -> sieve^#(_1), activate^#(n__filter(_0,_1)) -> filter^#(_0,_1), sieve^#(cons(_0,_1)) -> activate^#(_1), filter^#(s(s(_0)),cons(_1,_2)) -> activate^#(_2)]
  TRS = {primes -> sieve(from(s(s(0)))), from(_0) -> cons(_0,n__from(s(_0))), head(cons(_0,_1)) -> _0, tail(cons(_0,_1)) -> activate(_1), if(true,_0,_1) -> activate(_0), if(false,_0,_1) -> activate(_1), filter(s(s(_0)),cons(_1,_2)) -> if(divides(s(s(_0)),_1),n__filter(s(s(_0)),activate(_2)),n__cons(_1,n__filter(_0,sieve(_1)))), sieve(cons(_0,_1)) -> cons(_0,n__filter(_0,sieve(activate(_1)))), from(_0) -> n__from(_0), filter(_0,_1) -> n__filter(_0,_1), cons(_0,_1) -> n__cons(_0,_1), activate(n__from(_0)) -> from(_0), activate(n__filter(_0,_1)) -> filter(_0,_1), activate(n__cons(_0,_1)) -> cons(_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... Too many argument filtering possibilities (124416)! 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, 5 unfolded rules generated.
        # Iteration 1: no loop found, 14 unfolded rules generated.
        # Iteration 2: no loop found, 22 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 = sieve^#(cons(_0,_1)) -> sieve^#(activate(_1)) [trans] is in U_IR^0.
        D = sieve^#(cons(_0,_1)) -> activate^#(_1) is a dependency pair of IR.
        We build a composed triple from L0 and D.
        ==> L1 = [sieve^#(cons(_0,_1)) -> sieve^#(activate(_1)), sieve^#(cons(_2,_3)) -> activate^#(_3)] [comp] is in U_IR^1.
        Let p1 = [0].
        We unfold the first rule of L1 forwards at position p1
        with the rule activate(n__from(_0)) -> from(_0).
        ==> L2 = [sieve^#(cons(_0,n__from(_1))) -> sieve^#(from(_1)), sieve^#(cons(_2,_3)) -> activate^#(_3)] [comp] is in U_IR^2.
        Let p2 = [0].
        We unfold the first rule of L2 forwards at position p2
        with the rule from(_0) -> cons(_0,n__from(s(_0))).
        ==> L3 = [sieve^#(cons(_0,n__from(_1))) -> sieve^#(cons(_1,n__from(s(_1)))), sieve^#(cons(_2,_3)) -> activate^#(_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 = 86