/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] ap(ap(ap(foldr,ap(f,_0)),_1),ap(ap(cons,_0),nil)) -> ap(ap(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)),_1)
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {_0->foldr, _1->ap(ap(cons,foldr),nil)} and theta2 = {}.
We have r|p = ap(ap(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)),_1) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = ap(ap(ap(foldr,ap(f,foldr)),ap(ap(cons,foldr),nil)),ap(ap(cons,foldr),nil)) 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 = [ap^#(ap(f,_0),_0) -> ap^#(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)), ap^#(ap(f,_0),_0) -> ap^#(_0,ap(f,_0)), ap^#(ap(f,_0),_0) -> ap^#(ap(cons,_0),nil), ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(ap(_0,_2),ap(ap(ap(foldr,_0),_1),_3)), ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(_0,_2), ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(ap(ap(foldr,_0),_1),_3), ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(ap(foldr,_0),_1)]
  TRS = {ap(ap(f,_0),_0) -> ap(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)), ap(ap(ap(foldr,_0),_1),nil) -> _1, ap(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap(ap(_0,_2),ap(ap(ap(foldr,_0),_1),_3))}
    ## 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=false, max=-1)
      # max_depth=4, unfold_variables=false:
        # Iteration 0: no loop found, 7 unfolded rules generated.
        # Iteration 1: no loop found, 52 unfolded rules generated.
        # Iteration 2: no loop found, 71 unfolded rules generated.
        # Iteration 3: success, found a loop, 22 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(ap(_0,_2),ap(ap(ap(foldr,_0),_1),_3)) [trans] is in U_IR^0.
        D = ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(_0,_2) is a dependency pair of IR.
        We build a composed triple from L0 and D.
        ==> L1 = [ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(ap(_0,_2),ap(ap(ap(foldr,_0),_1),_3)), ap^#(ap(ap(foldr,_4),_5),ap(ap(cons,_6),_7)) -> ap^#(_4,_6)] [comp] is in U_IR^1.
        Let p1 = [1].
        We unfold the first rule of L1 forwards at position p1
        with the rule ap(ap(ap(foldr,_0),_1),nil) -> _1.
        ==> L2 = [ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),nil)) -> ap^#(ap(_0,_2),_1), ap^#(ap(ap(foldr,_3),_4),ap(ap(cons,_5),_6)) -> ap^#(_3,_5)] [comp] is in U_IR^2.
        Let p2 = [0].
        We unfold the first rule of L2 forwards at position p2
        with the rule ap(ap(f,_0),_0) -> ap(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)).
        ==> L3 = [ap^#(ap(ap(foldr,ap(f,_0)),_1),ap(ap(cons,_0),nil)) -> ap^#(ap(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)),_1), ap^#(ap(ap(foldr,_2),_3),ap(ap(cons,_4),_5)) -> ap^#(_2,_4)] [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 = 510