/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 = 7] active(f(active(c),mark(b),active(c))) -> active(f(active(c),mark(b),active(c)))
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {} and theta2 = {}.
We have r|p = active(f(active(c),mark(b),active(c))) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = active(f(active(c),mark(b),active(c))) 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...
## 2 initial DP problems to solve.
## First, we try to decompose these problems into smaller problems.
## Round 1 [2 DP problems]:
  ## DP problem:
  Dependency pairs = [active^#(f(a,b,_0)) -> mark^#(f(_0,_0,_0)), mark^#(f(_0,_1,_2)) -> active^#(f(_0,_1,mark(_2))), mark^#(f(_0,_1,_2)) -> mark^#(_2)]
  TRS = {active(f(a,b,_0)) -> mark(f(_0,_0,_0)), active(c) -> mark(a), active(c) -> mark(b), mark(f(_0,_1,_2)) -> active(f(_0,_1,mark(_2))), mark(a) -> active(a), mark(b) -> active(b), mark(c) -> active(c), f(mark(_0),_1,_2) -> f(_0,_1,_2), f(_0,mark(_1),_2) -> f(_0,_1,_2), f(_0,_1,mark(_2)) -> f(_0,_1,_2), f(active(_0),_1,_2) -> f(_0,_1,_2), f(_0,active(_1),_2) -> f(_0,_1,_2), f(_0,_1,active(_2)) -> f(_0,_1,_2)}
    ## 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.
  ## DP problem:
  Dependency pairs = [f^#(mark(_0),_1,_2) -> f^#(_0,_1,_2), f^#(_0,mark(_1),_2) -> f^#(_0,_1,_2), f^#(_0,_1,mark(_2)) -> f^#(_0,_1,_2), f^#(active(_0),_1,_2) -> f^#(_0,_1,_2), f^#(_0,active(_1),_2) -> f^#(_0,_1,_2), f^#(_0,_1,active(_2)) -> f^#(_0,_1,_2)]
  TRS = {active(f(a,b,_0)) -> mark(f(_0,_0,_0)), active(c) -> mark(a), active(c) -> mark(b), mark(f(_0,_1,_2)) -> active(f(_0,_1,mark(_2))), mark(a) -> active(a), mark(b) -> active(b), mark(c) -> active(c), f(mark(_0),_1,_2) -> f(_0,_1,_2), f(_0,mark(_1),_2) -> f(_0,_1,_2), f(_0,_1,mark(_2)) -> f(_0,_1,_2), f(active(_0),_1,_2) -> f(_0,_1,_2), f(_0,active(_1),_2) -> f(_0,_1,_2), f(_0,_1,active(_2)) -> f(_0,_1,_2)}
    ## Trying with homeomorphic embeddings... Success!
  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=true, max=20)
      # max_depth=20, unfold_variables=false:
        # Iteration 0: no loop found, 3 unfolded rules generated.
        # Iteration 1: no loop found, 8 unfolded rules generated.
        # Iteration 2: no loop found, 54 unfolded rules generated.
        # Iteration 3: no loop found, 534 unfolded rules generated.
        # Iteration 4: no loop found, 5972 unfolded rules generated.
        # Iteration 5: no loop found, 71470 unfolded rules generated.
        # Iteration 6: no loop found, 888619 unfolded rules generated.
        # Iteration 7: success, found a loop, 47165 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = active^#(f(a,b,_0)) -> mark^#(f(_0,_0,_0)) [trans] is in U_IR^0.
        D = mark^#(f(_0,_1,_2)) -> active^#(f(_0,_1,mark(_2))) is a dependency pair of IR.
        We build a composed triple from L0 and D.
        ==> L1 = active^#(f(a,b,_0)) -> active^#(f(_0,_0,mark(_0))) [trans] is in U_IR^1.
        We build a unit triple from L1.
        ==> L2 = active^#(f(a,b,_0)) -> active^#(f(_0,_0,mark(_0))) [unit] is in U_IR^2.
        Let p2 = [0].
        We unfold the rule of L2 forwards at position p2
        with the rule f(_0,_1,mark(_2)) -> f(_0,_1,_2).
        ==> L3 = active^#(f(a,b,_0)) -> active^#(f(_0,_0,_0)) [unit] is in U_IR^3.
        Let p3 = [0].
        We unfold the rule of L3 backwards at position p3
        with the rule f(mark(_0),_1,_2) -> f(_0,_1,_2).
        ==> L4 = active^#(f(mark(a),b,_0)) -> active^#(f(_0,_0,_0)) [unit] is in U_IR^4.
        Let p4 = [0, 0].
        We unfold the rule of L4 backwards at position p4
        with the rule active(c) -> mark(a).
        ==> L5 = active^#(f(active(c),b,active(c))) -> active^#(f(active(c),active(c),active(c))) [unit] is in U_IR^5.
        Let p5 = [0].
        We unfold the rule of L5 backwards at position p5
        with the rule f(_0,mark(_1),_2) -> f(_0,_1,_2).
        ==> L6 = active^#(f(active(c),mark(b),active(c))) -> active^#(f(active(c),active(c),active(c))) [unit] is in U_IR^6.
        Let p6 = [0, 1].
        We unfold the rule of L6 forwards at position p6
        with the rule active(c) -> mark(b).
        ==> L7 = active^#(f(active(c),mark(b),active(c))) -> active^#(f(active(c),mark(b),active(c))) [unit] is in U_IR^7.
  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 = 1422679