/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 = 6] adx(cons(_0,n__zeros)) -> adx(cons(n__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->n__0}.
We have r|p = adx(cons(n__0,n__zeros)) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = adx(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...
## 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 = [adx^#(cons(_0,_1)) -> incr^#(cons(activate(_0),n__adx(activate(_1)))), activate^#(n__adx(_0)) -> adx^#(activate(_0)), incr^#(cons(_0,_1)) -> activate^#(_0), activate^#(n__incr(_0)) -> incr^#(activate(_0)), activate^#(n__incr(_0)) -> activate^#(_0), activate^#(n__adx(_0)) -> activate^#(_0), adx^#(cons(_0,_1)) -> activate^#(_1), adx^#(cons(_0,_1)) -> activate^#(_0), incr^#(cons(_0,_1)) -> activate^#(_1)]
  TRS = {nats -> adx(zeros), zeros -> cons(n__0,n__zeros), incr(cons(_0,_1)) -> cons(n__s(activate(_0)),n__incr(activate(_1))), adx(cons(_0,_1)) -> incr(cons(activate(_0),n__adx(activate(_1)))), hd(cons(_0,_1)) -> activate(_0), tl(cons(_0,_1)) -> activate(_1), 0 -> n__0, zeros -> n__zeros, s(_0) -> n__s(_0), incr(_0) -> n__incr(_0), adx(_0) -> n__adx(_0), activate(n__0) -> 0, activate(n__zeros) -> zeros, activate(n__s(_0)) -> s(_0), activate(n__incr(_0)) -> incr(activate(_0)), activate(n__adx(_0)) -> adx(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... Failed!
    ## 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, 9 unfolded rules generated.
        # Iteration 1: no loop found, 53 unfolded rules generated.
        # Iteration 2: no loop found, 162 unfolded rules generated.
        # Iteration 3: no loop found, 647 unfolded rules generated.
        # Iteration 4: no loop found, 1756 unfolded rules generated.
        # Iteration 5: no loop found, 5357 unfolded rules generated.
        # Iteration 6: success, found a loop, 13395 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = adx^#(cons(_0,_1)) -> incr^#(cons(activate(_0),n__adx(activate(_1)))) [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 = adx^#(cons(_0,_1)) -> activate^#(n__adx(activate(_1))) [trans] is in U_IR^1.
        D = activate^#(n__adx(_0)) -> adx^#(activate(_0)) is a dependency pair of IR.
        We build a composed triple from L1 and D.
        ==> L2 = [adx^#(cons(_0,_1)) -> activate^#(n__adx(activate(_1))), activate^#(n__adx(_2)) -> adx^#(activate(_2))] [comp] is in U_IR^2.
        Let p2 = [0, 0].
        We unfold the first rule of L2 forwards at position p2
        with the rule activate(n__zeros) -> zeros.
        ==> L3 = [adx^#(cons(_0,n__zeros)) -> activate^#(n__adx(zeros)), activate^#(n__adx(_1)) -> adx^#(activate(_1))] [comp] is in U_IR^3.
        Let p3 = [0, 0].
        We unfold the first rule of L3 forwards at position p3
        with the rule zeros -> cons(n__0,n__zeros).
        ==> L4 = adx^#(cons(_0,n__zeros)) -> adx^#(activate(cons(n__0,n__zeros))) [trans] is in U_IR^4.
        D = adx^#(cons(_0,_1)) -> activate^#(_1) is a dependency pair of IR.
        We build a composed triple from L4 and D.
        ==> L5 = [adx^#(cons(_0,n__zeros)) -> adx^#(activate(cons(n__0,n__zeros))), adx^#(cons(_1,_2)) -> activate^#(_2)] [comp] is in U_IR^5.
        Let p5 = [0].
        We unfold the first rule of L5 forwards at position p5
        with the rule activate(_0) -> _0.
        ==> L6 = [adx^#(cons(_0,n__zeros)) -> adx^#(cons(n__0,n__zeros)), adx^#(cons(_1,_2)) -> activate^#(_2)] [comp] is in U_IR^6.
  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 = 45853