TRS Standard pair #615306043

details

property	value
status	complete
benchmark	double.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n055.star.cs.uiowa.edu
space	AProVE_10

run statistics

property	value
solver	NTI_23
configuration	default
runtime (wallclock)	0.398046970367 seconds
cpu usage	0.884370665
max memory	5.9408384E7

stage attributes

key	value
output-size	4081
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: f(tt,s(s(0))) from [Emmes, Enger, Giesl, IJCAR'12]
* Description:
The following pattern rule was generated by the strategy
presented in Sect. 3 of [Emmes, Enger, Giesl, IJCAR'12]:
[iteration = 2] f(tt,s(s(_0))){_0->s(s(_0))}^n{_0->0} -> f(tt,s(s(s(s(_0))))){_0->s(s(_0))}^n{_0->0}
We apply Theorem 8 of [Emmes, Enger, Giesl, IJCAR'12]
on this rule with m = 1, b = 1, pi = epsilon,
sigma' = {} and mu' = {}.
Hence the term f(tt,s(s(0))), obtained from
instantiating n with 0 in the left-hand side of
the rule, 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 = [f^#(tt,_0) -> f^#(isDouble(_0),s(s(_0)))]
  TRS = {f(tt,_0) -> f(isDouble(_0),s(s(_0))), isDouble(s(s(_0))) -> isDouble(_0), isDouble(0) -> tt}
    ## 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.
      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 = [isDouble^#(s(s(_0))) -> isDouble^#(_0)]
  TRS = {f(tt,_0) -> f(isDouble(_0),s(s(_0))), isDouble(s(s(_0))) -> isDouble(_0), isDouble(0) -> tt}
    ## 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: [Emmes, Enger, Giesl, IJCAR'12] (forward=true, backward=true, max=-1). 
      # Iteration 0: nontermination not proved, 1 unfolded rule generated.
      # Iteration 1: nontermination not proved, 5 unfolded rules generated.
      # Iteration 2: success, nontermination proved, 14 unfolded rules generated.
      Here is the successful unfolding. Let IR be the TRS under analysis.
      IR contains the dependency pair f^#(tt,_0) -> f^#(isDouble(_0),s(s(_0))).
      We apply (I) of [Emmes, Enger, Giesl, IJCAR'12] to this dependency pair.
      ==> P0 = f^#(tt,_0){}^n{} -> f^#(isDouble(_0),s(s(_0))){}^n{} is in U_IR^0.
      We apply (VI) of [Emmes, Enger, Giesl, IJCAR'12] to this pattern rule
      at position [0] using the pattern rule
      isDouble(s(s(_0))){_0->s(s(_0))}^n{_0->_1} -> isDouble(_1){_0->s(s(_0))}^n{_0->_1}
      obtained from IR.
      ==> P1 = f^#(tt,s(s(_0))){_0->s(s(_0))}^n{_0->_1} -> f^#(isDouble(_1),s(s(s(s(_0))))){_0->s(s(_0))}^n{_0->_1} is in U_IR^1.
      We apply (V) + (IX) of [Emmes, Enger, Giesl, IJCAR'12] to this pattern rule
      at position [0] using the rule
      isDouble(0) -> tt
      of IR.
      ==> P2 = f^#(tt,s(s(_0))){_0->s(s(_0))}^n{_0->0} -> f^#(tt,s(s(s(s(_0))))){_0->s(s(_0))}^n{_0->0} is in U_IR^2.
  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 = 50

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