TRS Standard pair #615300842

details

property	value
status	complete
benchmark	Ex1_GM99_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n058.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	NTI_23
configuration	default
runtime (wallclock)	0.449799060822 seconds
cpu usage	1.145696703
max memory	8.6499328E7

stage attributes

key	value
output-size	4358
starexec-result	"NO"

output

/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


NO

** BEGIN proof argument **
* Certificate: a__f(a__c,b,a__c) is non-terminating
* Description:
The following rule R was generated while unfolding
the dependency pairs of the analyzed TRS:
[iteration = 5] a__f^#(a__c,b,a__c) -> a__f^#(a__c,b,a__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 = a__f^#(a__c,b,a__c)
and theta2(theta1(l)) = theta1(r|p), i.e., l semi-unifies with r|p.
So, the term theta1(l) = a__f(a__c,b,a__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...
## 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 = [a__f^#(a,b,_0) -> a__f^#(_0,_0,mark(_0)), mark^#(f(_0,_1,_2)) -> a__f^#(_0,_1,mark(_2)), mark^#(f(_0,_1,_2)) -> mark^#(_2), a__f^#(a,b,_0) -> mark^#(_0)]
  TRS = {a__f(a,b,_0) -> a__f(_0,_0,mark(_0)), a__c -> a, a__c -> b, mark(f(_0,_1,_2)) -> a__f(_0,_1,mark(_2)), mark(c) -> a__c, mark(a) -> a, mark(b) -> b, a__f(_0,_1,_2) -> f(_0,_1,_2), a__c -> c}
    ## 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 (22 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.
## 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, 4 unfolded rules generated.
        # Iteration 1: infiniteness not proved, 12 unfolded rules generated.
        # Iteration 2: infiniteness not proved, 36 unfolded rules generated.
        # Iteration 3: infiniteness not proved, 111 unfolded rules generated.
        # Iteration 4: infiniteness not proved, 402 unfolded rules generated.
        # Iteration 5: success, loop found, 98 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = a__f^#(a,b,_0) -> a__f^#(_0,_0,mark(_0)) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = a__f^#(a,b,_0) -> a__f^#(_0,_0,mark(_0)) [unit] is in U_IR^1.
        Let p1 = [0].
        We unfold the rule of L1 backwards at position p1
        with the rule a__c -> a.
        ==> L2 = a__f^#(a__c,b,a__c) -> a__f^#(a__c,a__c,mark(a__c)) [unit] is in U_IR^2.
        Let p2 = [1].
        We unfold the rule of L2 forwards at position p2
        with the rule a__c -> b.
        ==> L3 = a__f^#(a__c,b,a__c) -> a__f^#(a__c,b,mark(a__c)) [unit] is in U_IR^3.
        Let p3 = [2, 0].
        We unfold the rule of L3 forwards at position p3
        with the rule a__c -> c.
        ==> L4 = a__f^#(a__c,b,a__c) -> a__f^#(a__c,b,mark(c)) [unit] is in U_IR^4.
        Let p4 = [2].
        We unfold the rule of L4 forwards at position p4
        with the rule mark(c) -> a__c.
        ==> L5 = a__f^#(a__c,b,a__c) -> a__f^#(a__c,b,a__c) [unit] is in U_IR^5.
  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 **
Total number of generated unfolded rules = 1328

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