TRS Standard pair #615299071

details
property	value
status	complete
benchmark	Ex1_GM99_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n175.star.cs.uiowa.edu
space	Transformed_CSR_04
run statistics
property	value
solver	NTI_23
configuration	default
runtime (wallclock)	25.6584999561 seconds
cpu usage	80.69159447
max memory	1.112283136E10
stage attributes
key	value
output-size	6358
starexec-result	"NO"
output
/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


NO

** BEGIN proof argument **
* Certificate: active(f(active(c),mark(b),active(c))) is non-terminating
* Description:
The following rule R was generated while unfolding
the dependency pairs of 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 R.
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), i.e., l semi-unifies with r|p.
So, the term theta1(l) = active(f(active(c),mark(b),active(c)))
starts an infinite rewrite sequence 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)}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Failed!
    ## DP Processor: lexicographic path orders. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Failed!
    ## DP Processor: polynomial interpretations. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      This DP problem is too complex (15 coefficients to instantiate)! Aborting!
      Failed!
    ## DP Processor: Knuth-Bendix orders. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Failed!
  All the DP processors have 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)}
    ## DP Processor: homeomorphic embeddings. 
      Using no argument filtering, i.e., applying the processor to the full problem only.
      Success, finiteness proved!
  This DP problem is finite.
## A DP problem could not be proved finite.
## Now, we try to prove that this problem is infinite.
    ## DP Processor: [Payet, LOPSTR'18] + [Payet, Non-termination in TRS and LP] (forward=true, backward=true, max=20). 
      # max_depth=20, unfold_variables=false:
        # Iteration 0: infiniteness not proved, 3 unfolded rules generated.
        # Iteration 1: infiniteness not proved, 8 unfolded rules generated.
        # Iteration 2: infiniteness not proved, 54 unfolded rules generated.
        # Iteration 3: infiniteness not proved, 534 unfolded rules generated.
        # Iteration 4: infiniteness not proved, 5972 unfolded rules generated.
        # Iteration 5: infiniteness not proved, 71470 unfolded rules generated.
        # Iteration 6: infiniteness not proved, 888619 unfolded rules generated.
        # Iteration 7: success, loop found, 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 3.10.0-1160.62.1.el7.x86_64 for amd64
using Java version 1.8.0_322
** END proof description **
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