We consider the system Applicative_first_order_05__#3.10.

  Alphabet:

    0 : [] --> b 
    add : [b * c] --> c 
    app : [c * c] --> c 
    eq : [b * b] --> a 
    false : [] --> a 
    filter : [b -> a * c] --> c 
    filter2 : [a * b -> a * b * c] --> c 
    if!fac6220min : [a * c] --> b 
    if!fac6220minsort : [a * c * c] --> c 
    if!fac6220rm : [a * b * c] --> c 
    le : [b * b] --> a 
    map : [b -> b * c] --> c 
    min : [c] --> b 
    minsort : [c * c] --> c 
    nil : [] --> c 
    rm : [b * c] --> c 
    s : [b] --> b 
    true : [] --> a 

  Rules:

    eq(0, 0) => true 
    eq(0, s(x)) => false 
    eq(s(x), 0) => false 
    eq(s(x), s(y)) => eq(x, y) 
    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    app(nil, x) => x 
    app(add(x, y), z) => add(x, app(y, z)) 
    min(add(x, nil)) => x 
    min(add(x, add(y, z))) => if!fac6220min(le(x, y), add(x, add(y, z))) 
    if!fac6220min(true, add(x, add(y, z))) => min(add(x, z)) 
    if!fac6220min(false, add(x, add(y, z))) => min(add(y, z)) 
    rm(x, nil) => nil 
    rm(x, add(y, z)) => if!fac6220rm(eq(x, y), x, add(y, z)) 
    if!fac6220rm(true, x, add(y, z)) => rm(x, z) 
    if!fac6220rm(false, x, add(y, z)) => add(y, rm(x, z)) 
    minsort(nil, nil) => nil 
    minsort(add(x, y), z) => if!fac6220minsort(eq(x, min(add(x, y))), add(x, y), z) 
    if!fac6220minsort(true, add(x, y), z) => add(x, minsort(app(rm(x, y), z), nil)) 
    if!fac6220minsort(false, add(x, y), z) => minsort(y, add(x, z)) 
    map(f, nil) => nil 
    map(f, add(x, y)) => add(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, add(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => add(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(f, y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [FuhKop19]).

We thus obtain the following dependency pair problem (P_0, R_0, computable, formative):

  Dependency Pairs P_0:

    0] eq#(s(X), s(Y)) =#> eq#(X, Y)   
    1] le#(s(X), s(Y)) =#> le#(X, Y)   
    2] app#(add(X, Y), Z) =#> app#(Y, Z)   
    3] min#(add(X, add(Y, Z))) =#> if!fac6220min#(le(X, Y), add(X, add(Y, Z)))   
    4] min#(add(X, add(Y, Z))) =#> le#(X, Y)   
    5] if!fac6220min#(true, add(X, add(Y, Z))) =#> min#(add(X, Z))   
    6] if!fac6220min#(false, add(X, add(Y, Z))) =#> min#(add(Y, Z))   
    7] rm#(X, add(Y, Z)) =#> if!fac6220rm#(eq(X, Y), X, add(Y, Z))   
    8] rm#(X, add(Y, Z)) =#> eq#(X, Y)   
    9] if!fac6220rm#(true, X, add(Y, Z)) =#> rm#(X, Z)   
    10] if!fac6220rm#(false, X, add(Y, Z)) =#> rm#(X, Z)   
    11] minsort#(add(X, Y), Z) =#> if!fac6220minsort#(eq(X, min(add(X, Y))), add(X, Y), Z)   
    12] minsort#(add(X, Y), Z) =#> eq#(X, min(add(X, Y)))   
    13] minsort#(add(X, Y), Z) =#> min#(add(X, Y))   
    14] if!fac6220minsort#(true, add(X, Y), Z) =#> minsort#(app(rm(X, Y), Z), nil)   
    15] if!fac6220minsort#(true, add(X, Y), Z) =#> app#(rm(X, Y), Z)   
    16] if!fac6220minsort#(true, add(X, Y), Z) =#> rm#(X, Y)   
    17] if!fac6220minsort#(false, add(X, Y), Z) =#> minsort#(Y, add(X, Z))   
    18] map#(F, add(X, Y)) =#> map#(F, Y)   
    19] filter#(F, add(X, Y)) =#> filter2#(F X, F, X, Y)   
    20] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    21] filter2#(false, F, X, Y) =#> filter#(F, 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) 
    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    app(nil, X) => X 
    app(add(X, Y), Z) => add(X, app(Y, Z)) 
    min(add(X, nil)) => X 
    min(add(X, add(Y, Z))) => if!fac6220min(le(X, Y), add(X, add(Y, Z))) 
    if!fac6220min(true, add(X, add(Y, Z))) => min(add(X, Z)) 
    if!fac6220min(false, add(X, add(Y, Z))) => min(add(Y, Z)) 
    rm(X, nil) => nil 
    rm(X, add(Y, Z)) => if!fac6220rm(eq(X, Y), X, add(Y, Z)) 
    if!fac6220rm(true, X, add(Y, Z)) => rm(X, Z) 
    if!fac6220rm(false, X, add(Y, Z)) => add(Y, rm(X, Z)) 
    minsort(nil, nil) => nil 
    minsort(add(X, Y), Z) => if!fac6220minsort(eq(X, min(add(X, Y))), add(X, Y), Z) 
    if!fac6220minsort(true, add(X, Y), Z) => add(X, minsort(app(rm(X, Y), Z), nil)) 
    if!fac6220minsort(false, add(X, Y), Z) => minsort(Y, add(X, Z)) 
    map(F, nil) => nil 
    map(F, add(X, Y)) => add(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, add(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => add(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(F, Y) 

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

We consider the dependency pair problem (P_0, R_0, computable, 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 : 2 
    * 3 : 5, 6 
    * 4 : 1 
    * 5 : 3, 4 
    * 6 : 3, 4 
    * 7 : 9, 10 
    * 8 : 0 
    * 9 : 7, 8 
    * 10 : 7, 8 
    * 11 : 14, 15, 16, 17 
    * 12 : 0 
    * 13 : 3, 4 
    * 14 : 11, 12, 13 
    * 15 : 2 
    * 16 : 7, 8 
    * 17 : 11, 12, 13 
    * 18 : 18 
    * 19 : 20, 21 
    * 20 : 19 
    * 21 : 19 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

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

  P_3:

    app#(add(X, Y), Z) =#> app#(Y, Z)   

  P_4:

    min#(add(X, add(Y, Z))) =#> if!fac6220min#(le(X, Y), add(X, add(Y, Z)))   
    if!fac6220min#(true, add(X, add(Y, Z))) =#> min#(add(X, Z))   
    if!fac6220min#(false, add(X, add(Y, Z))) =#> min#(add(Y, Z))   

  P_5:

    rm#(X, add(Y, Z)) =#> if!fac6220rm#(eq(X, Y), X, add(Y, Z))   
    if!fac6220rm#(true, X, add(Y, Z)) =#> rm#(X, Z)   
    if!fac6220rm#(false, X, add(Y, Z)) =#> rm#(X, Z)   

  P_6:

    minsort#(add(X, Y), Z) =#> if!fac6220minsort#(eq(X, min(add(X, Y))), add(X, Y), Z)   
    if!fac6220minsort#(true, add(X, Y), Z) =#> minsort#(app(rm(X, Y), Z), nil)   
    if!fac6220minsort#(false, add(X, Y), Z) =#> minsort#(Y, add(X, Z))   

  P_7:

    map#(F, add(X, Y)) =#> map#(F, Y)   

  P_8:

    filter#(F, add(X, Y)) =#> filter2#(F X, F, X, Y)   
    filter2#(true, F, X, Y) =#> filter#(F, Y)   
    filter2#(false, F, X, Y) =#> filter#(F, 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), (P_3, R_0, m, f), (P_4, R_0, m, f), (P_5, R_0, m, f), (P_6, R_0, m, f), (P_7, R_0, m, f) and (P_8, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative), (P_4, R_0, computable, formative), (P_5, R_0, computable, formative), (P_6, R_0, computable, formative), (P_7, R_0, computable, formative) and (P_8, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_8, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(filter2#) = 4 
  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, add(X, Y))) = add(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_8, R_0, computable, f) by (P_9, R_0, computable, f), where P_9 contains:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative), (P_4, R_0, computable, formative), (P_5, R_0, computable, formative), (P_6, R_0, computable, formative), (P_7, R_0, computable, formative) and (P_9, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_9, R_0, computable, 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 :  

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 each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative), (P_4, R_0, computable, formative), (P_5, R_0, computable, formative), (P_6, R_0, computable, formative) and (P_7, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_7, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(map#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(map#(F, add(X, Y))) = add(X, Y) |> Y = nu(map#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_7, R_0, computable, f) by ({}, R_0, computable, f).  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, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative), (P_4, R_0, computable, formative), (P_5, R_0, computable, formative) and (P_6, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_6, R_0, computable, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_6, R_0) are:

  eq(0, 0) => true 
  eq(0, s(X)) => false 
  eq(s(X), 0) => false 
  eq(s(X), s(Y)) => eq(X, Y) 
  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 
  app(nil, X) => X 
  app(add(X, Y), Z) => add(X, app(Y, Z)) 
  min(add(X, nil)) => X 
  min(add(X, add(Y, Z))) => if!fac6220min(le(X, Y), add(X, add(Y, Z))) 
  if!fac6220min(true, add(X, add(Y, Z))) => min(add(X, Z)) 
  if!fac6220min(false, add(X, add(Y, Z))) => min(add(Y, Z)) 
  rm(X, nil) => nil 
  rm(X, add(Y, Z)) => if!fac6220rm(eq(X, Y), X, add(Y, Z)) 
  if!fac6220rm(true, X, add(Y, Z)) => rm(X, Z) 
  if!fac6220rm(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:

  minsort#(add(X, Y), Z) >? if!fac6220minsort#(eq(X, min(add(X, Y))), add(X, Y), Z) 
  if!fac6220minsort#(true, add(X, Y), Z) >? minsort#(app(rm(X, Y), Z), nil) 
  if!fac6220minsort#(false, add(X, Y), Z) >? minsort#(Y, add(X, Z)) 
  eq(0, 0) >= true 
  eq(0, s(X)) >= false 
  eq(s(X), 0) >= false 
  eq(s(X), s(Y)) >= eq(X, Y) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  app(nil, X) >= X 
  app(add(X, Y), Z) >= add(X, app(Y, Z)) 
  min(add(X, nil)) >= X 
  min(add(X, add(Y, Z))) >= if!fac6220min(le(X, Y), add(X, add(Y, Z))) 
  if!fac6220min(true, add(X, add(Y, Z))) >= min(add(X, Z)) 
  if!fac6220min(false, add(X, add(Y, Z))) >= min(add(Y, Z)) 
  rm(X, nil) >= nil 
  rm(X, add(Y, Z)) >= if!fac6220rm(eq(X, Y), X, add(Y, Z)) 
  if!fac6220rm(true, X, add(Y, Z)) >= rm(X, Z) 
  if!fac6220rm(false, X, add(Y, Z)) >= add(Y, rm(X, Z)) 

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

The following interpretation satisfies the requirements:

  0 = 3 
  add = \y0y1.2 + y0 + y1 
  app = \y0y1.y0 + y1 
  eq = \y0y1.0 
  false = 0 
  if!fac6220min = \y0y1.2y1 
  if!fac6220minsort# = \y0y1y2.2y1 + 2y2 
  if!fac6220rm = \y0y1y2.y2 
  le = \y0y1.2 
  min = \y0.2y0 
  minsort# = \y0y1.2y0 + 2y1 
  nil = 0 
  rm = \y0y1.y1 
  s = \y0.3 
  true = 0 

Using this interpretation, the requirements translate to:

  [[minsort#(add(_x0, _x1), _x2)]] = 4 + 2x0 + 2x1 + 2x2 >= 4 + 2x0 + 2x1 + 2x2 = [[if!fac6220minsort#(eq(_x0, min(add(_x0, _x1))), add(_x0, _x1), _x2)]] 
  [[if!fac6220minsort#(true, add(_x0, _x1), _x2)]] = 4 + 2x0 + 2x1 + 2x2 > 2x1 + 2x2 = [[minsort#(app(rm(_x0, _x1), _x2), nil)]] 
  [[if!fac6220minsort#(false, add(_x0, _x1), _x2)]] = 4 + 2x0 + 2x1 + 2x2 >= 4 + 2x0 + 2x1 + 2x2 = [[minsort#(_x1, add(_x0, _x2))]] 
  [[eq(0, 0)]] = 0 >= 0 = [[true]] 
  [[eq(0, s(_x0))]] = 0 >= 0 = [[false]] 
  [[eq(s(_x0), 0)]] = 0 >= 0 = [[false]] 
  [[eq(s(_x0), s(_x1))]] = 0 >= 0 = [[eq(_x0, _x1)]] 
  [[le(0, _x0)]] = 2 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 2 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 2 >= 2 = [[le(_x0, _x1)]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(add(_x0, _x1), _x2)]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[add(_x0, app(_x1, _x2))]] 
  [[min(add(_x0, nil))]] = 4 + 2x0 >= x0 = [[_x0]] 
  [[min(add(_x0, add(_x1, _x2)))]] = 8 + 2x0 + 2x1 + 2x2 >= 8 + 2x0 + 2x1 + 2x2 = [[if!fac6220min(le(_x0, _x1), add(_x0, add(_x1, _x2)))]] 
  [[if!fac6220min(true, add(_x0, add(_x1, _x2)))]] = 8 + 2x0 + 2x1 + 2x2 >= 4 + 2x0 + 2x2 = [[min(add(_x0, _x2))]] 
  [[if!fac6220min(false, add(_x0, add(_x1, _x2)))]] = 8 + 2x0 + 2x1 + 2x2 >= 4 + 2x1 + 2x2 = [[min(add(_x1, _x2))]] 
  [[rm(_x0, nil)]] = 0 >= 0 = [[nil]] 
  [[rm(_x0, add(_x1, _x2))]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[if!fac6220rm(eq(_x0, _x1), _x0, add(_x1, _x2))]] 
  [[if!fac6220rm(true, _x0, add(_x1, _x2))]] = 2 + x1 + x2 >= x2 = [[rm(_x0, _x2)]] 
  [[if!fac6220rm(false, _x0, add(_x1, _x2))]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[add(_x1, rm(_x0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_6, R_0, computable, formative) by (P_10, R_0, computable, formative), where P_10 consists of:

  minsort#(add(X, Y), Z) =#> if!fac6220minsort#(eq(X, min(add(X, Y))), add(X, Y), Z)   
  if!fac6220minsort#(false, add(X, Y), Z) =#> minsort#(Y, add(X, Z))   

Thus, the original system is terminating if each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative), (P_4, R_0, computable, formative), (P_5, R_0, computable, formative) and (P_10, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_10, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(if!fac6220minsort#) = 2 
  nu(minsort#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(minsort#(add(X, Y), Z)) = add(X, Y) = add(X, Y) = nu(if!fac6220minsort#(eq(X, min(add(X, Y))), add(X, Y), Z)) 
  nu(if!fac6220minsort#(false, add(X, Y), Z)) = add(X, Y) |> Y = nu(minsort#(Y, add(X, Z))) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_10, R_0, computable, f) by (P_11, R_0, computable, f), where P_11 contains:

  minsort#(add(X, Y), Z) =#> if!fac6220minsort#(eq(X, min(add(X, Y))), add(X, Y), Z)   

Thus, the original system is terminating if each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative), (P_4, R_0, computable, formative), (P_5, R_0, computable, formative) and (P_11, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_11, R_0, computable, 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 each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative), (P_4, R_0, computable, formative) and (P_5, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_5, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(if!fac6220rm#) = 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(if!fac6220rm#(eq(X, Y), X, add(Y, Z))) 
  nu(if!fac6220rm#(true, X, add(Y, Z))) = add(Y, Z) |> Z = nu(rm#(X, Z)) 
  nu(if!fac6220rm#(false, X, add(Y, Z))) = add(Y, Z) |> Z = nu(rm#(X, Z)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_5, R_0, computable, f) by (P_12, R_0, computable, f), where P_12 contains:

  rm#(X, add(Y, Z)) =#> if!fac6220rm#(eq(X, Y), X, add(Y, Z))   

Thus, the original system is terminating if each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative), (P_4, R_0, computable, formative) and (P_12, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_12, R_0, computable, 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 each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative) and (P_4, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_4, R_0, computable, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_4, R_0) are:

  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 

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

  min#(add(X, add(Y, Z))) >? if!fac6220min#(le(X, Y), add(X, add(Y, Z))) 
  if!fac6220min#(true, add(X, add(Y, Z))) >? min#(add(X, Z)) 
  if!fac6220min#(false, add(X, add(Y, Z))) >? min#(add(Y, Z)) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 

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

The following interpretation satisfies the requirements:

  0 = 3 
  add = \y0y1.2 + 3y0 + 3y1 
  false = 0 
  if!fac6220min# = \y0y1.y1 
  le = \y0y1.y1 + 2y0 
  min# = \y0.1 + y0 
  s = \y0.3 + 2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[min#(add(_x0, add(_x1, _x2)))]] = 9 + 3x0 + 9x1 + 9x2 > 8 + 3x0 + 9x1 + 9x2 = [[if!fac6220min#(le(_x0, _x1), add(_x0, add(_x1, _x2)))]] 
  [[if!fac6220min#(true, add(_x0, add(_x1, _x2)))]] = 8 + 3x0 + 9x1 + 9x2 > 3 + 3x0 + 3x2 = [[min#(add(_x0, _x2))]] 
  [[if!fac6220min#(false, add(_x0, add(_x1, _x2)))]] = 8 + 3x0 + 9x1 + 9x2 > 3 + 3x1 + 3x2 = [[min#(add(_x1, _x2))]] 
  [[le(0, _x0)]] = 6 + x0 >= 0 = [[true]] 
  [[le(s(_x0), 0)]] = 9 + 4x0 >= 0 = [[false]] 
  [[le(s(_x0), s(_x1))]] = 9 + 2x1 + 4x0 >= x1 + 2x0 = [[le(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_0) by ({}, R_0).  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, computable, formative), (P_2, R_0, computable, formative) and (P_3, R_0, computable, formative) is finite.

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

We apply the subterm criterion with the following projection function:

  nu(app#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(app#(add(X, Y), Z)) = add(X, Y) |> Y = nu(app#(Y, Z)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_3, R_0, computable, f) by ({}, R_0, computable, f).  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, computable, formative) and (P_2, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_2, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(le#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, computable, f) by ({}, R_0, computable, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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

We consider the dependency pair problem (P_1, R_0, computable, 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 [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, computable, f) by ({}, R_0, computable, 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 +++

[FuhKop19]  C. Fuhs, and C. Kop.  A static higher-order dependency pair framework.  In Proceedings of ESOP 2019, 2019.
[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.