TRS Stand 20472 pair #381710916

details

property	value
status	complete
benchmark	quotminus.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n038.star.cs.uiowa.edu
space	Rubio_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.103343963623 seconds
cpu usage	0.092094362
max memory	4136960.0

stage attributes

key	value
output-size	5859
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 
  Z : [] --> o 
  min : [o * o] --> o 
  plus : [o * o] --> o 
  quot : [o * o] --> o 
  s : [o] --> o 

  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  min(X, 0) => X 
  min(s(X), s(Y)) => min(X, Y) 
  min(min(X, Y), Z) => min(X, plus(Y, Z)) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(min(X, Y), s(Y))) 

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] plus#(s(X), Y) =#> plus#(X, Y)   
    1] min#(s(X), s(Y)) =#> min#(X, Y)   
    2] min#(min(X, Y), Z) =#> min#(X, plus(Y, Z))   
    3] min#(min(X, Y), Z) =#> plus#(Y, Z)   
    4] quot#(s(X), s(Y)) =#> quot#(min(X, Y), s(Y))   
    5] quot#(s(X), s(Y)) =#> min#(X, Y)   

  Rules R_0:

    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    min(X, 0) => X 
    min(s(X), s(Y)) => min(X, Y) 
    min(min(X, Y), Z) => min(X, plus(Y, Z)) 
    quot(0, s(X)) => 0 
    quot(s(X), s(Y)) => s(quot(min(X, Y), s(Y))) 

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

This graph has the following strongly connected components:

  P_1:

    plus#(s(X), Y) =#> plus#(X, Y)   

  P_2:

    min#(s(X), s(Y)) =#> min#(X, Y)   
    min#(min(X, Y), Z) =#> min#(X, plus(Y, Z))   

  P_3:

    quot#(s(X), s(Y)) =#> quot#(min(X, Y), s(Y))   

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), (P_2, R_0, m, f) and (P_3, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_3, R_0) are:

  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  min(X, 0) => X 
  min(s(X), s(Y)) => min(X, Y) 
  min(min(X, Y), Z) => min(X, plus(Y, Z)) 

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  quot#(s(X), s(Y)) >? quot#(min(X, Y), s(Y)) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  min(X, 0) >= X 
  min(s(X), s(Y)) >= min(X, Y) 
  min(min(X, Y), Z) >= min(X, plus(Y, Z))

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