TRS Stand 20472 pair #381711069

details

property	value
status	complete
benchmark	t013.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n079.star.cs.uiowa.edu
space	HirokawaMiddeldorp_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0856690406799 seconds
cpu usage	0.08208732
max memory	3506176.0

stage attributes

key	value
output-size	4942
starexec-result	YES

output

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


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


YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  !minus : [o * o] --> o 
  0 : [] --> o 
  f : [o] --> o 
  g : [o] --> o 
  s : [o] --> o 

  !minus(X, 0) => X 
  !minus(0, s(X)) => 0 
  !minus(s(X), s(Y)) => !minus(X, Y) 
  f(0) => 0 
  f(s(X)) => !minus(s(X), g(f(X))) 
  g(0) => s(0) 
  g(s(X)) => !minus(s(X), f(g(X))) 

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]).

We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] !minus#(s(X), s(Y)) =#> !minus#(X, Y)   
    1] f#(s(X)) =#> !minus#(s(X), g(f(X)))   
    2] f#(s(X)) =#> g#(f(X))   
    3] f#(s(X)) =#> f#(X)   
    4] g#(s(X)) =#> !minus#(s(X), f(g(X)))   
    5] g#(s(X)) =#> f#(g(X))   
    6] g#(s(X)) =#> g#(X)   

  Rules R_0:

    !minus(X, 0) => X 
    !minus(0, s(X)) => 0 
    !minus(s(X), s(Y)) => !minus(X, Y) 
    f(0) => 0 
    f(s(X)) => !minus(s(X), g(f(X))) 
    g(0) => s(0) 
    g(s(X)) => !minus(s(X), f(g(X))) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 0 
    * 1 : 0 
    * 2 : 4, 5, 6 
    * 3 : 1, 2, 3 
    * 4 : 0 
    * 5 : 1, 2, 3 
    * 6 : 4, 5, 6 

This graph has the following strongly connected components:

  P_1:

    !minus#(s(X), s(Y)) =#> !minus#(X, Y)   

  P_2:

    f#(s(X)) =#> g#(f(X))   
    f#(s(X)) =#> f#(X)   
    g#(s(X)) =#> f#(g(X))   
    g#(s(X)) =#> g#(X)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_0, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  f#(s(X)) >? g#(f(X)) 
  f#(s(X)) >? f#(X) 
  g#(s(X)) >? f#(g(X)) 
  g#(s(X)) >? g#(X) 
  !minus(X, 0) >= X 
  !minus(0, s(X)) >= 0 
  !minus(s(X), s(Y)) >= !minus(X, Y) 
  f(0) >= 0 
  f(s(X)) >= !minus(s(X), g(f(X))) 
  g(0) >= s(0) 
  g(s(X)) >= !minus(s(X), f(g(X))) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

popout

output may be truncated. 'popout' for the full output.

job log

popout

actions all output return to TRS Stand 20472