TRS Stand 20472 pair #381710081

details

property	value
status	complete
benchmark	12.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n104.star.cs.uiowa.edu
space	Various_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.130774974823 seconds
cpu usage	0.127327127
max memory	5509120.0

stage attributes

key	value
output-size	7290
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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.

  !plus : [o * o] --> o 
  !times : [o * o] --> o 
  0 : [] --> o 
  I : [o] --> o 
  O : [o] --> o 

  O(0) => 0 
  !plus(0, X) => X 
  !plus(X, 0) => X 
  !plus(O(X), O(Y)) => O(!plus(X, Y)) 
  !plus(O(X), I(Y)) => I(!plus(X, Y)) 
  !plus(I(X), O(Y)) => I(!plus(X, Y)) 
  !plus(I(X), I(Y)) => O(!plus(!plus(X, Y), I(0))) 
  !times(0, X) => 0 
  !times(X, 0) => 0 
  !times(O(X), Y) => O(!times(X, Y)) 
  !times(I(X), Y) => !plus(O(!times(X, Y)), 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#(O(X), O(Y)) =#> O#(!plus(X, Y))   
    1] !plus#(O(X), O(Y)) =#> !plus#(X, Y)   
    2] !plus#(O(X), I(Y)) =#> !plus#(X, Y)   
    3] !plus#(I(X), O(Y)) =#> !plus#(X, Y)   
    4] !plus#(I(X), I(Y)) =#> O#(!plus(!plus(X, Y), I(0)))   
    5] !plus#(I(X), I(Y)) =#> !plus#(!plus(X, Y), I(0))   
    6] !plus#(I(X), I(Y)) =#> !plus#(X, Y)   
    7] !times#(O(X), Y) =#> O#(!times(X, Y))   
    8] !times#(O(X), Y) =#> !times#(X, Y)   
    9] !times#(I(X), Y) =#> !plus#(O(!times(X, Y)), Y)   
    10] !times#(I(X), Y) =#> O#(!times(X, Y))   
    11] !times#(I(X), Y) =#> !times#(X, Y)   

  Rules R_0:

    O(0) => 0 
    !plus(0, X) => X 
    !plus(X, 0) => X 
    !plus(O(X), O(Y)) => O(!plus(X, Y)) 
    !plus(O(X), I(Y)) => I(!plus(X, Y)) 
    !plus(I(X), O(Y)) => I(!plus(X, Y)) 
    !plus(I(X), I(Y)) => O(!plus(!plus(X, Y), I(0))) 
    !times(0, X) => 0 
    !times(X, 0) => 0 
    !times(O(X), Y) => O(!times(X, Y)) 
    !times(I(X), Y) => !plus(O(!times(X, Y)), 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 :  
    * 1 : 0, 1, 2, 3, 4, 5, 6 
    * 2 : 0, 1, 2, 3, 4, 5, 6 
    * 3 : 0, 1, 2, 3, 4, 5, 6 
    * 4 :  
    * 5 : 2, 4, 5, 6 
    * 6 : 0, 1, 2, 3, 4, 5, 6 
    * 7 :  
    * 8 : 7, 8, 9, 10, 11 
    * 9 : 0, 1, 2 
    * 10 :  
    * 11 : 7, 8, 9, 10, 11 

This graph has the following strongly connected components:

  P_1:

    !plus#(O(X), O(Y)) =#> !plus#(X, Y)   
    !plus#(O(X), I(Y)) =#> !plus#(X, Y)   
    !plus#(I(X), O(Y)) =#> !plus#(X, Y)   
    !plus#(I(X), I(Y)) =#> !plus#(!plus(X, Y), I(0))   
    !plus#(I(X), I(Y)) =#> !plus#(X, Y)   

  P_2:

    !times#(O(X), Y) =#> !times#(X, Y)   
    !times#(I(X), Y) =#> !times#(X, 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) 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.

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