TRS Stand 20472 pair #381712269

details

property	value
status	complete
benchmark	secret5.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n033.star.cs.uiowa.edu
space	Secret_07_TRS

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.328936815262 seconds
cpu usage	0.291583126
max memory	1.0186752E7

stage attributes

key	value
output-size	14143
starexec-result	YES

output

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


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YES
We consider the system theBenchmark.

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

  0 : [] --> o 
  a : [o] --> o 
  cs : [o * o] --> o 
  d : [o] --> o 
  f : [o * o] --> o 
  nf : [o * o] --> o 
  nil : [] --> o 
  ns : [o] --> o 
  nt : [o] --> o 
  p : [o * o] --> o 
  q : [o] --> o 
  r : [o] --> o 
  s : [o] --> o 
  t : [o] --> o 

  t(X) => cs(r(q(X)), nt(ns(X))) 
  q(0) => 0 
  q(s(X)) => s(p(q(X), d(X))) 
  d(0) => 0 
  d(s(X)) => s(s(d(X))) 
  p(0, X) => X 
  p(X, 0) => X 
  p(s(X), s(Y)) => s(s(p(X, Y))) 
  f(0, X) => nil 
  f(s(X), cs(Y, Z)) => cs(Y, nf(X, a(Z))) 
  t(X) => nt(X) 
  s(X) => ns(X) 
  f(X, Y) => nf(X, Y) 
  a(nt(X)) => t(a(X)) 
  a(ns(X)) => s(a(X)) 
  a(nf(X, Y)) => f(a(X), a(Y)) 
  a(X) => 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] t#(X) =#> q#(X)   
    1] q#(s(X)) =#> s#(p(q(X), d(X)))   
    2] q#(s(X)) =#> p#(q(X), d(X))   
    3] q#(s(X)) =#> q#(X)   
    4] q#(s(X)) =#> d#(X)   
    5] d#(s(X)) =#> s#(s(d(X)))   
    6] d#(s(X)) =#> s#(d(X))   
    7] d#(s(X)) =#> d#(X)   
    8] p#(s(X), s(Y)) =#> s#(s(p(X, Y)))   
    9] p#(s(X), s(Y)) =#> s#(p(X, Y))   
    10] p#(s(X), s(Y)) =#> p#(X, Y)   
    11] f#(s(X), cs(Y, Z)) =#> a#(Z)   
    12] a#(nt(X)) =#> t#(a(X))   
    13] a#(nt(X)) =#> a#(X)   
    14] a#(ns(X)) =#> s#(a(X))   
    15] a#(ns(X)) =#> a#(X)   
    16] a#(nf(X, Y)) =#> f#(a(X), a(Y))   
    17] a#(nf(X, Y)) =#> a#(X)   
    18] a#(nf(X, Y)) =#> a#(Y)   

  Rules R_0:

    t(X) => cs(r(q(X)), nt(ns(X))) 
    q(0) => 0 
    q(s(X)) => s(p(q(X), d(X))) 
    d(0) => 0 
    d(s(X)) => s(s(d(X))) 
    p(0, X) => X 
    p(X, 0) => X 
    p(s(X), s(Y)) => s(s(p(X, Y))) 
    f(0, X) => nil 
    f(s(X), cs(Y, Z)) => cs(Y, nf(X, a(Z))) 
    t(X) => nt(X) 
    s(X) => ns(X) 
    f(X, Y) => nf(X, Y) 
    a(nt(X)) => t(a(X)) 
    a(ns(X)) => s(a(X)) 
    a(nf(X, Y)) => f(a(X), a(Y)) 
    a(X) => 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 : 1, 2, 3, 4 
    * 1 :  
    * 2 : 8, 9, 10 
    * 3 : 1, 2, 3, 4

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