TRS Stand 20472 pair #381716300

details

property	value
status	complete
benchmark	Ex4_7_15_Bor03_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n046.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.42076086998 seconds
cpu usage	0.399469558
max memory	1.071104E7

stage attributes

key	value
output-size	17304
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 
  active : [o] --> o 
  cons : [o * o] --> o 
  f : [o] --> o 
  mark : [o] --> o 
  p : [o] --> o 
  s : [o] --> o 

  active(f(0)) => mark(cons(0, f(s(0)))) 
  active(f(s(0))) => mark(f(p(s(0)))) 
  active(p(s(0))) => mark(0) 
  mark(f(X)) => active(f(mark(X))) 
  mark(0) => active(0) 
  mark(cons(X, Y)) => active(cons(mark(X), Y)) 
  mark(s(X)) => active(s(mark(X))) 
  mark(p(X)) => active(p(mark(X))) 
  f(mark(X)) => f(X) 
  f(active(X)) => f(X) 
  cons(mark(X), Y) => cons(X, Y) 
  cons(X, mark(Y)) => cons(X, Y) 
  cons(active(X), Y) => cons(X, Y) 
  cons(X, active(Y)) => cons(X, Y) 
  s(mark(X)) => s(X) 
  s(active(X)) => s(X) 
  p(mark(X)) => p(X) 
  p(active(X)) => p(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] active#(f(0)) =#> mark#(cons(0, f(s(0))))   
    1] active#(f(0)) =#> cons#(0, f(s(0)))   
    2] active#(f(0)) =#> f#(s(0))   
    3] active#(f(0)) =#> s#(0)   
    4] active#(f(s(0))) =#> mark#(f(p(s(0))))   
    5] active#(f(s(0))) =#> f#(p(s(0)))   
    6] active#(f(s(0))) =#> p#(s(0))   
    7] active#(f(s(0))) =#> s#(0)   
    8] active#(p(s(0))) =#> mark#(0)   
    9] mark#(f(X)) =#> active#(f(mark(X)))   
    10] mark#(f(X)) =#> f#(mark(X))   
    11] mark#(f(X)) =#> mark#(X)   
    12] mark#(0) =#> active#(0)   
    13] mark#(cons(X, Y)) =#> active#(cons(mark(X), Y))   
    14] mark#(cons(X, Y)) =#> cons#(mark(X), Y)   
    15] mark#(cons(X, Y)) =#> mark#(X)   
    16] mark#(s(X)) =#> active#(s(mark(X)))   
    17] mark#(s(X)) =#> s#(mark(X))   
    18] mark#(s(X)) =#> mark#(X)   
    19] mark#(p(X)) =#> active#(p(mark(X)))   
    20] mark#(p(X)) =#> p#(mark(X))   
    21] mark#(p(X)) =#> mark#(X)   
    22] f#(mark(X)) =#> f#(X)   
    23] f#(active(X)) =#> f#(X)   
    24] cons#(mark(X), Y) =#> cons#(X, Y)   
    25] cons#(X, mark(Y)) =#> cons#(X, Y)   
    26] cons#(active(X), Y) =#> cons#(X, Y)   
    27] cons#(X, active(Y)) =#> cons#(X, Y)   
    28] s#(mark(X)) =#> s#(X)   
    29] s#(active(X)) =#> s#(X)   
    30] p#(mark(X)) =#> p#(X)   
    31] p#(active(X)) =#> p#(X)   

  Rules R_0:

    active(f(0)) => mark(cons(0, f(s(0)))) 
    active(f(s(0))) => mark(f(p(s(0)))) 
    active(p(s(0))) => mark(0) 
    mark(f(X)) => active(f(mark(X))) 
    mark(0) => active(0) 
    mark(cons(X, Y)) => active(cons(mark(X), Y)) 
    mark(s(X)) => active(s(mark(X))) 
    mark(p(X)) => active(p(mark(X))) 
    f(mark(X)) => f(X) 
    f(active(X)) => f(X) 
    cons(mark(X), Y) => cons(X, Y) 
    cons(X, mark(Y)) => cons(X, Y) 
    cons(active(X), Y) => cons(X, Y) 
    cons(X, active(Y)) => cons(X, Y) 
    s(mark(X)) => s(X) 
    s(active(X)) => s(X) 
    p(mark(X)) => p(X) 
    p(active(X)) => p(X) 

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

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