/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 = 4] quote1(n__cons(_0,n__from(_1))) -> quote1(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 = quote1(n__cons(activate(_1),n__from(n__s(activate(_1))))) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = quote1(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...
## 6 initial DP problems to solve.
## First, we try to decompose these problems into smaller problems.
## Round 1 [6 DP problems]:
  ## DP problem:
  Dependency pairs = [unquote1^#(cons1(_0,_1)) -> unquote1^#(_1)]
  TRS = {sel(s(_0),cons(_1,_2)) -> sel(_0,activate(_2)), sel(0,cons(_0,_1)) -> _0, first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,n__first(_0,activate(_2))), from(_0) -> cons(_0,n__from(n__s(_0))), sel1(s(_0),cons(_1,_2)) -> sel1(_0,activate(_2)), sel1(0,cons(_0,_1)) -> quote(_0), first1(0,_0) -> nil1, first1(s(_0),cons(_1,_2)) -> cons1(quote(_1),first1(_0,activate(_2))), quote(n__0) -> 01, quote1(n__cons(_0,_1)) -> cons1(quote(activate(_0)),quote1(activate(_1))), quote1(n__nil) -> nil1, quote(n__s(_0)) -> s1(quote(activate(_0))), quote(n__sel(_0,_1)) -> sel1(activate(_0),activate(_1)), quote1(n__first(_0,_1)) -> first1(activate(_0),activate(_1)), unquote(01) -> 0, unquote(s1(_0)) -> s(unquote(_0)), unquote1(nil1) -> nil, unquote1(cons1(_0,_1)) -> fcons(unquote(_0),unquote1(_1)), fcons(_0,_1) -> cons(_0,_1), first(_0,_1) -> n__first(_0,_1), from(_0) -> n__from(_0), s(_0) -> n__s(_0), 0 -> n__0, cons(_0,_1) -> n__cons(_0,_1), nil -> n__nil, sel(_0,_1) -> n__sel(_0,_1), activate(n__first(_0,_1)) -> first(activate(_0),activate(_1)), activate(n__from(_0)) -> from(activate(_0)), activate(n__s(_0)) -> s(activate(_0)), activate(n__0) -> 0, activate(n__cons(_0,_1)) -> cons(activate(_0),_1), activate(n__nil) -> nil, activate(n__sel(_0,_1)) -> sel(activate(_0),activate(_1)), activate(_0) -> _0}
    ## Trying with homeomorphic embeddings... Success!
  This DP problem is finite.
  ## DP problem:
  Dependency pairs = [unquote^#(s1(_0)) -> unquote^#(_0)]
  TRS = {sel(s(_0),cons(_1,_2)) -> sel(_0,activate(_2)), sel(0,cons(_0,_1)) -> _0, first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,n__first(_0,activate(_2))), from(_0) -> cons(_0,n__from(n__s(_0))), sel1(s(_0),cons(_1,_2)) -> sel1(_0,activate(_2)), sel1(0,cons(_0,_1)) -> quote(_0), first1(0,_0) -> nil1, first1(s(_0),cons(_1,_2)) -> cons1(quote(_1),first1(_0,activate(_2))), quote(n__0) -> 01, quote1(n__cons(_0,_1)) -> cons1(quote(activate(_0)),quote1(activate(_1))), quote1(n__nil) -> nil1, quote(n__s(_0)) -> s1(quote(activate(_0))), quote(n__sel(_0,_1)) -> sel1(activate(_0),activate(_1)), quote1(n__first(_0,_1)) -> first1(activate(_0),activate(_1)), unquote(01) -> 0, unquote(s1(_0)) -> s(unquote(_0)), unquote1(nil1) -> nil, unquote1(cons1(_0,_1)) -> fcons(unquote(_0),unquote1(_1)), fcons(_0,_1) -> cons(_0,_1), first(_0,_1) -> n__first(_0,_1), from(_0) -> n__from(_0), s(_0) -> n__s(_0), 0 -> n__0, cons(_0,_1) -> n__cons(_0,_1), nil -> n__nil, sel(_0,_1) -> n__sel(_0,_1), activate(n__first(_0,_1)) -> first(activate(_0),activate(_1)), activate(n__from(_0)) -> from(activate(_0)), activate(n__s(_0)) -> s(activate(_0)), activate(n__0) -> 0, activate(n__cons(_0,_1)) -> cons(activate(_0),_1), activate(n__nil) -> nil, activate(n__sel(_0,_1)) -> sel(activate(_0),activate(_1)), activate(_0) -> _0}
    ## Trying with homeomorphic embeddings... Success!
  This DP problem is finite.
  ## DP problem:
  Dependency pairs = [quote1^#(n__cons(_0,_1)) -> quote1^#(activate(_1))]
  TRS = {sel(s(_0),cons(_1,_2)) -> sel(_0,activate(_2)), sel(0,cons(_0,_1)) -> _0, first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,n__first(_0,activate(_2))), from(_0) -> cons(_0,n__from(n__s(_0))), sel1(s(_0),cons(_1,_2)) -> sel1(_0,activate(_2)), sel1(0,cons(_0,_1)) -> quote(_0), first1(0,_0) -> nil1, first1(s(_0),cons(_1,_2)) -> cons1(quote(_1),first1(_0,activate(_2))), quote(n__0) -> 01, quote1(n__cons(_0,_1)) -> cons1(quote(activate(_0)),quote1(activate(_1))), quote1(n__nil) -> nil1, quote(n__s(_0)) -> s1(quote(activate(_0))), quote(n__sel(_0,_1)) -> sel1(activate(_0),activate(_1)), quote1(n__first(_0,_1)) -> first1(activate(_0),activate(_1)), unquote(01) -> 0, unquote(s1(_0)) -> s(unquote(_0)), unquote1(nil1) -> nil, unquote1(cons1(_0,_1)) -> fcons(unquote(_0),unquote1(_1)), fcons(_0,_1) -> cons(_0,_1), first(_0,_1) -> n__first(_0,_1), from(_0) -> n__from(_0), s(_0) -> n__s(_0), 0 -> n__0, cons(_0,_1) -> n__cons(_0,_1), nil -> n__nil, sel(_0,_1) -> n__sel(_0,_1), activate(n__first(_0,_1)) -> first(activate(_0),activate(_1)), activate(n__from(_0)) -> from(activate(_0)), activate(n__s(_0)) -> s(activate(_0)), activate(n__0) -> 0, activate(n__cons(_0,_1)) -> cons(activate(_0),_1), activate(n__nil) -> nil, activate(n__sel(_0,_1)) -> sel(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... Too many argument filtering possibilities (60466176)! Aborting!
    ## Trying with Knuth-Bendix orders... This DP problem is too complex! Aborting!
  Don't know whether this DP problem is finite.
  ## DP problem:
  Dependency pairs = [first1^#(s(_0),cons(_1,_2)) -> first1^#(_0,activate(_2))]
  TRS = {sel(s(_0),cons(_1,_2)) -> sel(_0,activate(_2)), sel(0,cons(_0,_1)) -> _0, first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,n__first(_0,activate(_2))), from(_0) -> cons(_0,n__from(n__s(_0))), sel1(s(_0),cons(_1,_2)) -> sel1(_0,activate(_2)), sel1(0,cons(_0,_1)) -> quote(_0), first1(0,_0) -> nil1, first1(s(_0),cons(_1,_2)) -> cons1(quote(_1),first1(_0,activate(_2))), quote(n__0) -> 01, quote1(n__cons(_0,_1)) -> cons1(quote(activate(_0)),quote1(activate(_1))), quote1(n__nil) -> nil1, quote(n__s(_0)) -> s1(quote(activate(_0))), quote(n__sel(_0,_1)) -> sel1(activate(_0),activate(_1)), quote1(n__first(_0,_1)) -> first1(activate(_0),activate(_1)), unquote(01) -> 0, unquote(s1(_0)) -> s(unquote(_0)), unquote1(nil1) -> nil, unquote1(cons1(_0,_1)) -> fcons(unquote(_0),unquote1(_1)), fcons(_0,_1) -> cons(_0,_1), first(_0,_1) -> n__first(_0,_1), from(_0) -> n__from(_0), s(_0) -> n__s(_0), 0 -> n__0, cons(_0,_1) -> n__cons(_0,_1), nil -> n__nil, sel(_0,_1) -> n__sel(_0,_1), activate(n__first(_0,_1)) -> first(activate(_0),activate(_1)), activate(n__from(_0)) -> from(activate(_0)), activate(n__s(_0)) -> s(activate(_0)), activate(n__0) -> 0, activate(n__cons(_0,_1)) -> cons(activate(_0),_1), activate(n__nil) -> nil, activate(n__sel(_0,_1)) -> sel(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... Too many argument filtering possibilities (60466176)! Aborting!
    ## Trying with Knuth-Bendix orders... This DP problem is too complex! Aborting!
  Don't know whether this DP problem is finite.
  ## DP problem:
  Dependency pairs = [sel1^#(s(_0),cons(_1,_2)) -> sel1^#(_0,activate(_2)), quote^#(n__sel(_0,_1)) -> sel1^#(activate(_0),activate(_1)), quote^#(n__s(_0)) -> quote^#(activate(_0)), sel1^#(0,cons(_0,_1)) -> quote^#(_0)]
  TRS = {sel(s(_0),cons(_1,_2)) -> sel(_0,activate(_2)), sel(0,cons(_0,_1)) -> _0, first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,n__first(_0,activate(_2))), from(_0) -> cons(_0,n__from(n__s(_0))), sel1(s(_0),cons(_1,_2)) -> sel1(_0,activate(_2)), sel1(0,cons(_0,_1)) -> quote(_0), first1(0,_0) -> nil1, first1(s(_0),cons(_1,_2)) -> cons1(quote(_1),first1(_0,activate(_2))), quote(n__0) -> 01, quote1(n__cons(_0,_1)) -> cons1(quote(activate(_0)),quote1(activate(_1))), quote1(n__nil) -> nil1, quote(n__s(_0)) -> s1(quote(activate(_0))), quote(n__sel(_0,_1)) -> sel1(activate(_0),activate(_1)), quote1(n__first(_0,_1)) -> first1(activate(_0),activate(_1)), unquote(01) -> 0, unquote(s1(_0)) -> s(unquote(_0)), unquote1(nil1) -> nil, unquote1(cons1(_0,_1)) -> fcons(unquote(_0),unquote1(_1)), fcons(_0,_1) -> cons(_0,_1), first(_0,_1) -> n__first(_0,_1), from(_0) -> n__from(_0), s(_0) -> n__s(_0), 0 -> n__0, cons(_0,_1) -> n__cons(_0,_1), nil -> n__nil, sel(_0,_1) -> n__sel(_0,_1), activate(n__first(_0,_1)) -> first(activate(_0),activate(_1)), activate(n__from(_0)) -> from(activate(_0)), activate(n__s(_0)) -> s(activate(_0)), activate(n__0) -> 0, activate(n__cons(_0,_1)) -> cons(activate(_0),_1), activate(n__nil) -> nil, activate(n__sel(_0,_1)) -> sel(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... Too many argument filtering possibilities (60466176)! Aborting!
    ## Trying with Knuth-Bendix orders... This DP problem is too complex! Aborting!
  Don't know whether this DP problem is finite.
  ## DP problem:
  Dependency pairs = [sel^#(s(_0),cons(_1,_2)) -> sel^#(_0,activate(_2)), activate^#(n__sel(_0,_1)) -> sel^#(activate(_0),activate(_1)), first^#(s(_0),cons(_1,_2)) -> activate^#(_2), activate^#(n__first(_0,_1)) -> first^#(activate(_0),activate(_1)), activate^#(n__first(_0,_1)) -> activate^#(_0), activate^#(n__first(_0,_1)) -> activate^#(_1), activate^#(n__from(_0)) -> activate^#(_0), activate^#(n__s(_0)) -> activate^#(_0), activate^#(n__cons(_0,_1)) -> activate^#(_0), activate^#(n__sel(_0,_1)) -> activate^#(_0), activate^#(n__sel(_0,_1)) -> activate^#(_1), sel^#(s(_0),cons(_1,_2)) -> activate^#(_2)]
  TRS = {sel(s(_0),cons(_1,_2)) -> sel(_0,activate(_2)), sel(0,cons(_0,_1)) -> _0, first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,n__first(_0,activate(_2))), from(_0) -> cons(_0,n__from(n__s(_0))), sel1(s(_0),cons(_1,_2)) -> sel1(_0,activate(_2)), sel1(0,cons(_0,_1)) -> quote(_0), first1(0,_0) -> nil1, first1(s(_0),cons(_1,_2)) -> cons1(quote(_1),first1(_0,activate(_2))), quote(n__0) -> 01, quote1(n__cons(_0,_1)) -> cons1(quote(activate(_0)),quote1(activate(_1))), quote1(n__nil) -> nil1, quote(n__s(_0)) -> s1(quote(activate(_0))), quote(n__sel(_0,_1)) -> sel1(activate(_0),activate(_1)), quote1(n__first(_0,_1)) -> first1(activate(_0),activate(_1)), unquote(01) -> 0, unquote(s1(_0)) -> s(unquote(_0)), unquote1(nil1) -> nil, unquote1(cons1(_0,_1)) -> fcons(unquote(_0),unquote1(_1)), fcons(_0,_1) -> cons(_0,_1), first(_0,_1) -> n__first(_0,_1), from(_0) -> n__from(_0), s(_0) -> n__s(_0), 0 -> n__0, cons(_0,_1) -> n__cons(_0,_1), nil -> n__nil, sel(_0,_1) -> n__sel(_0,_1), activate(n__first(_0,_1)) -> first(activate(_0),activate(_1)), activate(n__from(_0)) -> from(activate(_0)), activate(n__s(_0)) -> s(activate(_0)), activate(n__0) -> 0, activate(n__cons(_0,_1)) -> cons(activate(_0),_1), activate(n__nil) -> nil, activate(n__sel(_0,_1)) -> sel(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... Too many argument filtering possibilities (60466176)! Aborting!
    ## Trying with Knuth-Bendix orders... This DP problem is too complex! Aborting!
  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=true, max=20)
      # max_depth=20, 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, 9 unfolded rules generated.
        # Iteration 3: no loop found, 102 unfolded rules generated.
        # Iteration 4: success, found a loop, 500 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = quote1^#(n__cons(_0,_1)) -> quote1^#(activate(_1)) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = quote1^#(n__cons(_0,_1)) -> quote1^#(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__from(_0)) -> from(activate(_0)).
        ==> L2 = quote1^#(n__cons(_0,n__from(_1))) -> quote1^#(from(activate(_1))) [unit] is in U_IR^2.
        Let p2 = [0].
        We unfold the rule of L2 forwards at position p2
        with the rule from(_0) -> cons(_0,n__from(n__s(_0))).
        ==> L3 = quote1^#(n__cons(_0,n__from(_1))) -> quote1^#(cons(activate(_1),n__from(n__s(activate(_1))))) [unit] is in U_IR^3.
        Let p3 = [0].
        We unfold the rule of L3 forwards at position p3
        with the rule cons(_0,_1) -> n__cons(_0,_1).
        ==> L4 = quote1^#(n__cons(_0,n__from(_1))) -> quote1^#(n__cons(activate(_1),n__from(n__s(activate(_1))))) [unit] 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 = 6495