/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] adx(cons(_0,n__zeros)) -> adx(cons(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->0}.
We have r|p = adx(cons(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 = [incr^#(cons(_0,_1)) -> activate^#(_1), activate^#(n__incr(_0)) -> incr^#(_0), adx^#(cons(_0,_1)) -> incr^#(cons(_0,n__adx(activate(_1)))), activate^#(n__adx(_0)) -> adx^#(_0), adx^#(cons(_0,_1)) -> activate^#(_1)]
  TRS = {incr(nil) -> nil, incr(cons(_0,_1)) -> cons(s(_0),n__incr(activate(_1))), adx(nil) -> nil, adx(cons(_0,_1)) -> incr(cons(_0,n__adx(activate(_1)))), nats -> adx(zeros), zeros -> cons(0,n__zeros), head(cons(_0,_1)) -> _0, tail(cons(_0,_1)) -> activate(_1), incr(_0) -> n__incr(_0), adx(_0) -> n__adx(_0), zeros -> n__zeros, activate(n__incr(_0)) -> incr(_0), activate(n__adx(_0)) -> adx(_0), activate(n__zeros) -> zeros, 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... 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, 5 unfolded rules generated.
        # Iteration 1: no loop found, 16 unfolded rules generated.
        # Iteration 2: no loop found, 26 unfolded rules generated.
        # Iteration 3: no loop found, 57 unfolded rules generated.
        # Iteration 4: success, found a loop, 37 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = adx^#(cons(_0,_1)) -> incr^#(cons(_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)) -> incr^#(cons(_0,n__adx(activate(_1)))), incr^#(cons(_2,_3)) -> activate^#(_3)] [comp] is in U_IR^1.
        Let p1 = [0, 1, 0].
        We unfold the first rule of L1 forwards at position p1
        with the rule activate(n__zeros) -> zeros.
        ==> L2 = [adx^#(cons(_0,n__zeros)) -> incr^#(cons(_0,n__adx(zeros))), incr^#(cons(_1,_2)) -> activate^#(_2)] [comp] is in U_IR^2.
        Let p2 = [0, 1, 0].
        We unfold the first rule of L2 forwards at position p2
        with the rule zeros -> cons(0,n__zeros).
        ==> L3 = adx^#(cons(_0,n__zeros)) -> activate^#(n__adx(cons(0,n__zeros))) [trans] is in U_IR^3.
        D = activate^#(n__adx(_0)) -> adx^#(_0) is a dependency pair of IR.
        We build a composed triple from L3 and D.
        ==> L4 = adx^#(cons(_0,n__zeros)) -> adx^#(cons(0,n__zeros)) [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 = 362