/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.

  0 : [] --> o 
  add : [o * o] --> o 
  eq : [o * o] --> o 
  false : [] --> o 
  ifrm : [o * o * o] --> o 
  nil : [] --> o 
  purge : [o] --> o 
  rm : [o * o] --> o 
  s : [o] --> o 
  true : [] --> o 

  eq(0, 0) => true 
  eq(0, s(X)) => false 
  eq(s(X), 0) => false 
  eq(s(X), s(Y)) => eq(X, Y) 
  rm(X, nil) => nil 
  rm(X, add(Y, Z)) => ifrm(eq(X, Y), X, add(Y, Z)) 
  ifrm(true, X, add(Y, Z)) => rm(X, Z) 
  ifrm(false, X, add(Y, Z)) => add(Y, rm(X, Z)) 
  purge(nil) => nil 
  purge(add(X, Y)) => add(X, purge(rm(X, Y))) 

As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95].  Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows:

  0 : [] --> ad 
  add : [ad * ad] --> ad 
  eq : [ad * ad] --> jb 
  false : [] --> jb 
  ifrm : [jb * ad * ad] --> ad 
  nil : [] --> ad 
  purge : [ad] --> ad 
  rm : [ad * ad] --> ad 
  s : [ad] --> ad 
  true : [] --> jb 

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] eq#(s(X), s(Y)) =#> eq#(X, Y)   
    1] rm#(X, add(Y, Z)) =#> ifrm#(eq(X, Y), X, add(Y, Z))   
    2] rm#(X, add(Y, Z)) =#> eq#(X, Y)   
    3] ifrm#(true, X, add(Y, Z)) =#> rm#(X, Z)   
    4] ifrm#(false, X, add(Y, Z)) =#> rm#(X, Z)   
    5] purge#(add(X, Y)) =#> purge#(rm(X, Y))   
    6] purge#(add(X, Y)) =#> rm#(X, Y)   

  Rules R_0:

    eq(0, 0) => true 
    eq(0, s(X)) => false 
    eq(s(X), 0) => false 
    eq(s(X), s(Y)) => eq(X, Y) 
    rm(X, nil) => nil 
    rm(X, add(Y, Z)) => ifrm(eq(X, Y), X, add(Y, Z)) 
    ifrm(true, X, add(Y, Z)) => rm(X, Z) 
    ifrm(false, X, add(Y, Z)) => add(Y, rm(X, Z)) 
    purge(nil) => nil 
    purge(add(X, Y)) => add(X, purge(rm(X, 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 : 3, 4 
    * 2 : 0 
    * 3 : 1, 2 
    * 4 : 1, 2 
    * 5 : 5, 6 
    * 6 : 1, 2 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

    rm#(X, add(Y, Z)) =#> ifrm#(eq(X, Y), X, add(Y, Z))   
    ifrm#(true, X, add(Y, Z)) =#> rm#(X, Z)   
    ifrm#(false, X, add(Y, Z)) =#> rm#(X, Z)   

  P_3:

    purge#(add(X, Y)) =#> purge#(rm(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), (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).

The formative rules of (P_3, R_0) are R_1 ::=

  eq(0, 0) => true 
  eq(0, s(X)) => false 
  eq(s(X), 0) => false 
  eq(s(X), s(Y)) => eq(X, Y) 
  rm(X, add(Y, Z)) => ifrm(eq(X, Y), X, add(Y, Z)) 
  ifrm(true, X, add(Y, Z)) => rm(X, Z) 
  ifrm(false, X, add(Y, Z)) => add(Y, rm(X, Z)) 
  purge(add(X, Y)) => add(X, purge(rm(X, Y))) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_3, R_0, minimal, formative) by (P_3, R_1, minimal, formative).

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_1, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_1, minimal, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_3, R_1) are:

  eq(0, 0) => true 
  eq(0, s(X)) => false 
  eq(s(X), 0) => false 
  eq(s(X), s(Y)) => eq(X, Y) 
  rm(X, add(Y, Z)) => ifrm(eq(X, Y), X, add(Y, Z)) 
  ifrm(true, X, add(Y, Z)) => rm(X, Z) 
  ifrm(false, X, add(Y, Z)) => add(Y, rm(X, Z)) 

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

  purge#(add(X, Y)) >? purge#(rm(X, Y)) 
  eq(0, 0) >= true 
  eq(0, s(X)) >= false 
  eq(s(X), 0) >= false 
  eq(s(X), s(Y)) >= eq(X, Y) 
  rm(X, add(Y, Z)) >= ifrm(eq(X, Y), X, add(Y, Z)) 
  ifrm(true, X, add(Y, Z)) >= rm(X, Z) 
  ifrm(false, X, add(Y, Z)) >= add(Y, rm(X, Z)) 

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

We consider usable_rules with respect to the following argument filtering:

  ifrm(x_1,x_2,x_3) = ifrm(x_2x_3) 

This leaves the following ordering requirements:

  purge#(add(X, Y)) > purge#(rm(X, Y)) 
  rm(X, add(Y, Z)) >= ifrm(eq(X, Y), X, add(Y, Z)) 
  ifrm(true, X, add(Y, Z)) >= rm(X, Z) 
  ifrm(false, X, add(Y, Z)) >= add(Y, rm(X, Z)) 

The following interpretation satisfies the requirements:

  0 = 3 
  add = \y0y1.3 + y1 
  eq = \y0y1.0 
  false = 0 
  ifrm = \y0y1y2.y2 
  purge# = \y0.3y0 
  rm = \y0y1.y1 
  s = \y0.3 
  true = 0 

Using this interpretation, the requirements translate to:

  [[purge#(add(_x0, _x1))]] = 9 + 3x1 > 3x1 = [[purge#(rm(_x0, _x1))]] 
  [[rm(_x0, add(_x1, _x2))]] = 3 + x2 >= 3 + x2 = [[ifrm(eq(_x0, _x1), _x0, add(_x1, _x2))]] 
  [[ifrm(true, _x0, add(_x1, _x2))]] = 3 + x2 >= x2 = [[rm(_x0, _x2)]] 
  [[ifrm(false, _x0, add(_x1, _x2))]] = 3 + x2 >= 3 + x2 = [[add(_x1, rm(_x0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_1) by ({}, R_1).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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 apply the subterm criterion with the following projection function:

  nu(ifrm#) = 3 
  nu(rm#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(rm#(X, add(Y, Z))) = add(Y, Z) = add(Y, Z) = nu(ifrm#(eq(X, Y), X, add(Y, Z))) 
  nu(ifrm#(true, X, add(Y, Z))) = add(Y, Z) |> Z = nu(rm#(X, Z)) 
  nu(ifrm#(false, X, add(Y, Z))) = add(Y, Z) |> Z = nu(rm#(X, Z)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_4, R_0, minimal, f), where P_4 contains:

  rm#(X, add(Y, Z)) =#> ifrm#(eq(X, Y), X, add(Y, Z))   

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_4, 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 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

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

We consider the dependency pair problem (P_1, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(eq#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(eq#(s(X), s(Y))) = s(X) |> X = nu(eq#(X, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[FuhGieParSchSwi11]  C. Fuhs, J. Giesl, M. Parting, P. Schneider-Kamp, and S. Swiderski.  Proving Termination by Dependency Pairs and Inductive Theorem Proving.  In volume 47(2) of Journal of Automated Reasoning.  133--160, 2011.
[Gra95]  B. Gramlich.  Abstract Relations Between Restricted Termination and Confluence Properties of Rewrite Systems.  In volume 24(1-2) of Fundamentae Informaticae.  3--23, 1995.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.