We consider the system 06plusmult.

  Alphabet:

    mult : [N * N] --> N 
    plus : [N * N] --> N 
    s : [N] --> N 
    z : [] --> N 

  Rules:

    plus(z, x) => x 
    plus(s(x), y) => plus(x, s(y)) 
    plus(plus(x, y), u) => plus(x, plus(y, u)) 
    mult(z, x) => z 
    mult(s(x), y) => plus(mult(x, y), y) 
    mult(plus(x, y), u) => plus(mult(x, u), mult(y, u)) 

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 [Kop13]).

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

  Dependency Pairs P_0:

    0] plus#(s(X), Y) =#> plus#(X, s(Y))   
    1] plus#(plus(X, Y), Z) =#> plus#(X, plus(Y, Z))   
    2] plus#(plus(X, Y), Z) =#> plus#(Y, Z)   
    3] mult#(s(X), Y) =#> plus#(mult(X, Y), Y)   
    4] mult#(s(X), Y) =#> mult#(X, Y)   
    5] mult#(plus(X, Y), Z) =#> plus#(mult(X, Z), mult(Y, Z))   
    6] mult#(plus(X, Y), Z) =#> mult#(X, Z)   
    7] mult#(plus(X, Y), Z) =#> mult#(Y, Z)   

  Rules R_0:

    plus(z, X) => X 
    plus(s(X), Y) => plus(X, s(Y)) 
    plus(plus(X, Y), Z) => plus(X, plus(Y, Z)) 
    mult(z, X) => z 
    mult(s(X), Y) => plus(mult(X, Y), Y) 
    mult(plus(X, Y), Z) => plus(mult(X, Z), mult(Y, Z)) 

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

We consider the dependency pair problem (P_0, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  plus#(s(X), Y) >? plus#(X, s(Y)) 
  plus#(plus(X, Y), Z) >? plus#(X, plus(Y, Z)) 
  plus#(plus(X, Y), Z) >? plus#(Y, Z) 
  mult#(s(X), Y) >? plus#(mult(X, Y), Y) 
  mult#(s(X), Y) >? mult#(X, Y) 
  mult#(plus(X, Y), Z) >? plus#(mult(X, Z), mult(Y, Z)) 
  mult#(plus(X, Y), Z) >? mult#(X, Z) 
  mult#(plus(X, Y), Z) >? mult#(Y, Z) 
  plus(z, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(z, X) >= z 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

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

The following interpretation satisfies the requirements:

  mult = \y0y1.2 + y1 
  mult# = \y0y1.3 + 2y1 
  plus = \y0y1.y1 
  plus# = \y0y1.0 
  s = \y0.0 
  z = 0 

Using this interpretation, the requirements translate to:

  [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, s(_x1))]] 
  [[plus#(plus(_x0, _x1), _x2)]] = 0 >= 0 = [[plus#(_x0, plus(_x1, _x2))]] 
  [[plus#(plus(_x0, _x1), _x2)]] = 0 >= 0 = [[plus#(_x1, _x2)]] 
  [[mult#(s(_x0), _x1)]] = 3 + 2x1 > 0 = [[plus#(mult(_x0, _x1), _x1)]] 
  [[mult#(s(_x0), _x1)]] = 3 + 2x1 >= 3 + 2x1 = [[mult#(_x0, _x1)]] 
  [[mult#(plus(_x0, _x1), _x2)]] = 3 + 2x2 > 0 = [[plus#(mult(_x0, _x2), mult(_x1, _x2))]] 
  [[mult#(plus(_x0, _x1), _x2)]] = 3 + 2x2 >= 3 + 2x2 = [[mult#(_x0, _x2)]] 
  [[mult#(plus(_x0, _x1), _x2)]] = 3 + 2x2 >= 3 + 2x2 = [[mult#(_x1, _x2)]] 
  [[plus(z, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = x1 >= 0 = [[plus(_x0, s(_x1))]] 
  [[plus(plus(_x0, _x1), _x2)]] = x2 >= x2 = [[plus(_x0, plus(_x1, _x2))]] 
  [[mult(z, _x0)]] = 2 + x0 >= 0 = [[z]] 
  [[mult(s(_x0), _x1)]] = 2 + x1 >= x1 = [[plus(mult(_x0, _x1), _x1)]] 
  [[mult(plus(_x0, _x1), _x2)]] = 2 + x2 >= 2 + x2 = [[plus(mult(_x0, _x2), mult(_x1, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of:

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

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

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  plus#(s(X), Y) >? plus#(X, s(Y)) 
  plus#(plus(X, Y), Z) >? plus#(X, plus(Y, Z)) 
  plus#(plus(X, Y), Z) >? plus#(Y, Z) 
  mult#(s(X), Y) >? mult#(X, Y) 
  mult#(plus(X, Y), Z) >? mult#(X, Z) 
  mult#(plus(X, Y), Z) >? mult#(Y, Z) 
  plus(z, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(z, X) >= z 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[mult#(x_1, x_2)]] = mult#(x_1) 
  [[plus#(x_1, x_2)]] = plus#(x_1) 
  [[z]] = _|_ 

We choose Lex = {mult, plus} and Mul = {mult#, plus#, s}, and the following precedence: mult > plus > mult# = plus# = s

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  plus#(s(X)) > plus#(X) 
  plus#(plus(X, Y)) >= plus#(X) 
  plus#(plus(X, Y)) >= plus#(Y) 
  mult#(s(X)) >= mult#(X) 
  mult#(plus(X, Y)) >= mult#(X) 
  mult#(plus(X, Y)) >= mult#(Y) 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(_|_, X) >= _|_ 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

With these choices, we have:

  1] plus#(s(X)) > plus#(X)  because [2], by definition 
  2] plus#*(s(X)) >= plus#(X)  because [3], by (Select) 
  3] s(X) >= plus#(X)  because s = plus#, s in Mul and [4], by (Fun) 
  4] X >= X  by (Meta) 

  5] plus#(plus(X, Y)) >= plus#(X)  because plus# in Mul and [6], by (Fun) 
  6] plus(X, Y) >= X  because [7], by (Star) 
  7] plus*(X, Y) >= X  because [8], by (Select) 
  8] X >= X  by (Meta) 

  9] plus#(plus(X, Y)) >= plus#(Y)  because plus# in Mul and [10], by (Fun) 
  10] plus(X, Y) >= Y  because [11], by (Star) 
  11] plus*(X, Y) >= Y  because [12], by (Select) 
  12] Y >= Y  by (Meta) 

  13] mult#(s(X)) >= mult#(X)  because [14], by (Star) 
  14] mult#*(s(X)) >= mult#(X)  because [15], by (Select) 
  15] s(X) >= mult#(X)  because s = mult#, s in Mul and [16], by (Fun) 
  16] X >= X  by (Meta) 

  17] mult#(plus(X, Y)) >= mult#(X)  because mult# in Mul and [18], by (Fun) 
  18] plus(X, Y) >= X  because [19], by (Star) 
  19] plus*(X, Y) >= X  because [20], by (Select) 
  20] X >= X  by (Meta) 

  21] mult#(plus(X, Y)) >= mult#(Y)  because mult# in Mul and [22], by (Fun) 
  22] plus(X, Y) >= Y  because [23], by (Star) 
  23] plus*(X, Y) >= Y  because [24], by (Select) 
  24] Y >= Y  by (Meta) 

  25] plus(_|_, X) >= X  because [26], by (Star) 
  26] plus*(_|_, X) >= X  because [27], by (Select) 
  27] X >= X  by (Meta) 

  28] plus(s(X), Y) >= plus(X, s(Y))  because [29], by (Star) 
  29] plus*(s(X), Y) >= plus(X, s(Y))  because [30], [32] and [34], by (Stat) 
  30] s(X) > X  because [31], by definition 
  31] s*(X) >= X  because [4], by (Select) 
  32] plus*(s(X), Y) >= X  because [33], by (Select) 
  33] s(X) >= X  because [31], by (Star) 
  34] plus*(s(X), Y) >= s(Y)  because plus > s and [35], by (Copy) 
  35] plus*(s(X), Y) >= Y  because [36], by (Select) 
  36] Y >= Y  by (Meta) 

  37] plus(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [38], by (Star) 
  38] plus*(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [39], [41] and [42], by (Stat) 
  39] plus(X, Y) > X  because [40], by definition 
  40] plus*(X, Y) >= X  because [8], by (Select) 
  41] plus*(plus(X, Y), Z) >= X  because [6], by (Select) 
  42] plus*(plus(X, Y), Z) >= plus(Y, Z)  because [43], [45] and [46], by (Stat) 
  43] plus(X, Y) > Y  because [44], by definition 
  44] plus*(X, Y) >= Y  because [12], by (Select) 
  45] plus*(plus(X, Y), Z) >= Y  because [10], by (Select) 
  46] plus*(plus(X, Y), Z) >= Z  because [47], by (Select) 
  47] Z >= Z  by (Meta) 

  48] mult(_|_, X) >= _|_  by (Bot) 

  49] mult(s(X), Y) >= plus(mult(X, Y), Y)  because [50], by (Star) 
  50] mult*(s(X), Y) >= plus(mult(X, Y), Y)  because mult > plus, [51] and [56], by (Copy) 
  51] mult*(s(X), Y) >= mult(X, Y)  because [52], [54] and [56], by (Stat) 
  52] s(X) > X  because [53], by definition 
  53] s*(X) >= X  because [16], by (Select) 
  54] mult*(s(X), Y) >= X  because [55], by (Select) 
  55] s(X) >= X  because [53], by (Star) 
  56] mult*(s(X), Y) >= Y  because [57], by (Select) 
  57] Y >= Y  by (Meta) 

  58] mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because [59], by (Star) 
  59] mult*(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because mult > plus, [60] and [66], by (Copy) 
  60] mult*(plus(X, Y), Z) >= mult(X, Z)  because [61], [63] and [64], by (Stat) 
  61] plus(X, Y) > X  because [62], by definition 
  62] plus*(X, Y) >= X  because [20], by (Select) 
  63] mult*(plus(X, Y), Z) >= X  because [18], by (Select) 
  64] mult*(plus(X, Y), Z) >= Z  because [65], by (Select) 
  65] Z >= Z  by (Meta) 
  66] mult*(plus(X, Y), Z) >= mult(Y, Z)  because [67], [69] and [64], by (Stat) 
  67] plus(X, Y) > Y  because [68], by definition 
  68] plus*(X, Y) >= Y  because [24], by (Select) 
  69] mult*(plus(X, Y), Z) >= Y  because [22], by (Select) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of:

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

Thus, the original system is terminating if (P_2, R_0, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  plus#(plus(X, Y), Z) >? plus#(X, plus(Y, Z)) 
  plus#(plus(X, Y), Z) >? plus#(Y, Z) 
  mult#(s(X), Y) >? mult#(X, Y) 
  mult#(plus(X, Y), Z) >? mult#(X, Z) 
  mult#(plus(X, Y), Z) >? mult#(Y, Z) 
  plus(z, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(z, X) >= z 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[plus#(x_1, x_2)]] = plus#(x_1) 
  [[z]] = _|_ 

We choose Lex = {plus} and Mul = {mult, mult#, plus#, s}, and the following precedence: mult# > plus# > mult > plus > s

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  plus#(plus(X, Y)) >= plus#(X) 
  plus#(plus(X, Y)) >= plus#(Y) 
  mult#(s(X), Y) >= mult#(X, Y) 
  mult#(plus(X, Y), Z) >= mult#(X, Z) 
  mult#(plus(X, Y), Z) > mult#(Y, Z) 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(_|_, X) >= _|_ 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

With these choices, we have:

  1] plus#(plus(X, Y)) >= plus#(X)  because plus# in Mul and [2], by (Fun) 
  2] plus(X, Y) >= X  because [3], by (Star) 
  3] plus*(X, Y) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

  5] plus#(plus(X, Y)) >= plus#(Y)  because plus# in Mul and [6], by (Fun) 
  6] plus(X, Y) >= Y  because [7], by (Star) 
  7] plus*(X, Y) >= Y  because [8], by (Select) 
  8] Y >= Y  by (Meta) 

  9] mult#(s(X), Y) >= mult#(X, Y)  because [10], by (Star) 
  10] mult#*(s(X), Y) >= mult#(X, Y)  because mult# in Mul, [11] and [14], by (Stat) 
  11] s(X) > X  because [12], by definition 
  12] s*(X) >= X  because [13], by (Select) 
  13] X >= X  by (Meta) 
  14] Y >= Y  by (Meta) 

  15] mult#(plus(X, Y), Z) >= mult#(X, Z)  because mult# in Mul, [16] and [19], by (Fun) 
  16] plus(X, Y) >= X  because [17], by (Star) 
  17] plus*(X, Y) >= X  because [18], by (Select) 
  18] X >= X  by (Meta) 
  19] Z >= Z  by (Meta) 

  20] mult#(plus(X, Y), Z) > mult#(Y, Z)  because [21], by definition 
  21] mult#*(plus(X, Y), Z) >= mult#(Y, Z)  because mult# in Mul, [22] and [19], by (Stat) 
  22] plus(X, Y) > Y  because [23], by definition 
  23] plus*(X, Y) >= Y  because [24], by (Select) 
  24] Y >= Y  by (Meta) 

  25] plus(_|_, X) >= X  because [26], by (Star) 
  26] plus*(_|_, X) >= X  because [27], by (Select) 
  27] X >= X  by (Meta) 

  28] plus(s(X), Y) >= plus(X, s(Y))  because [29], by (Star) 
  29] plus*(s(X), Y) >= plus(X, s(Y))  because [30], [33] and [35], by (Stat) 
  30] s(X) > X  because [31], by definition 
  31] s*(X) >= X  because [32], by (Select) 
  32] X >= X  by (Meta) 
  33] plus*(s(X), Y) >= X  because [34], by (Select) 
  34] s(X) >= X  because [31], by (Star) 
  35] plus*(s(X), Y) >= s(Y)  because plus > s and [36], by (Copy) 
  36] plus*(s(X), Y) >= Y  because [37], by (Select) 
  37] Y >= Y  by (Meta) 

  38] plus(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [39], by (Star) 
  39] plus*(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [40], [42] and [43], by (Stat) 
  40] plus(X, Y) > X  because [41], by definition 
  41] plus*(X, Y) >= X  because [4], by (Select) 
  42] plus*(plus(X, Y), Z) >= X  because [2], by (Select) 
  43] plus*(plus(X, Y), Z) >= plus(Y, Z)  because [44], [46] and [47], by (Stat) 
  44] plus(X, Y) > Y  because [45], by definition 
  45] plus*(X, Y) >= Y  because [8], by (Select) 
  46] plus*(plus(X, Y), Z) >= Y  because [6], by (Select) 
  47] plus*(plus(X, Y), Z) >= Z  because [48], by (Select) 
  48] Z >= Z  by (Meta) 

  49] mult(_|_, X) >= _|_  by (Bot) 

  50] mult(s(X), Y) >= plus(mult(X, Y), Y)  because [51], by (Star) 
  51] mult*(s(X), Y) >= plus(mult(X, Y), Y)  because mult > plus, [52] and [53], by (Copy) 
  52] mult*(s(X), Y) >= mult(X, Y)  because mult in Mul, [11] and [14], by (Stat) 
  53] mult*(s(X), Y) >= Y  because [14], by (Select) 

  54] mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because [55], by (Star) 
  55] mult*(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because mult > plus, [56] and [59], by (Copy) 
  56] mult*(plus(X, Y), Z) >= mult(X, Z)  because mult in Mul, [57] and [19], by (Stat) 
  57] plus(X, Y) > X  because [58], by definition 
  58] plus*(X, Y) >= X  because [18], by (Select) 
  59] mult*(plus(X, Y), Z) >= mult(Y, Z)  because mult in Mul, [22] and [19], by (Stat) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_0, static, formative) by (P_3, R_0, static, formative), where P_3 consists of:

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

Thus, the original system is terminating if (P_3, R_0, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  plus#(plus(X, Y), Z) >? plus#(X, plus(Y, Z)) 
  plus#(plus(X, Y), Z) >? plus#(Y, Z) 
  mult#(s(X), Y) >? mult#(X, Y) 
  mult#(plus(X, Y), Z) >? mult#(X, Z) 
  plus(z, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(z, X) >= z 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[mult#(x_1, x_2)]] = mult#(x_1) 
  [[plus#(x_1, x_2)]] = plus#(x_1) 
  [[z]] = _|_ 

We choose Lex = {plus} and Mul = {mult, mult#, plus#, s}, and the following precedence: mult > plus > mult# > plus# > s

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  plus#(plus(X, Y)) >= plus#(X) 
  plus#(plus(X, Y)) >= plus#(Y) 
  mult#(s(X)) >= mult#(X) 
  mult#(plus(X, Y)) > mult#(X) 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(_|_, X) >= _|_ 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

With these choices, we have:

  1] plus#(plus(X, Y)) >= plus#(X)  because [2], by (Star) 
  2] plus#*(plus(X, Y)) >= plus#(X)  because [3], by (Select) 
  3] plus(X, Y) >= plus#(X)  because [4], by (Star) 
  4] plus*(X, Y) >= plus#(X)  because plus > plus# and [5], by (Copy) 
  5] plus*(X, Y) >= X  because [6], by (Select) 
  6] X >= X  by (Meta) 

  7] plus#(plus(X, Y)) >= plus#(Y)  because plus# in Mul and [8], by (Fun) 
  8] plus(X, Y) >= Y  because [9], by (Star) 
  9] plus*(X, Y) >= Y  because [10], by (Select) 
  10] Y >= Y  by (Meta) 

  11] mult#(s(X)) >= mult#(X)  because mult# in Mul and [12], by (Fun) 
  12] s(X) >= X  because [13], by (Star) 
  13] s*(X) >= X  because [14], by (Select) 
  14] X >= X  by (Meta) 

  15] mult#(plus(X, Y)) > mult#(X)  because [16], by definition 
  16] mult#*(plus(X, Y)) >= mult#(X)  because [17], by (Select) 
  17] plus(X, Y) >= mult#(X)  because [18], by (Star) 
  18] plus*(X, Y) >= mult#(X)  because plus > mult# and [19], by (Copy) 
  19] plus*(X, Y) >= X  because [20], by (Select) 
  20] X >= X  by (Meta) 

  21] plus(_|_, X) >= X  because [22], by (Star) 
  22] plus*(_|_, X) >= X  because [23], by (Select) 
  23] X >= X  by (Meta) 

  24] plus(s(X), Y) >= plus(X, s(Y))  because [25], by (Star) 
  25] plus*(s(X), Y) >= plus(X, s(Y))  because [26], [29] and [31], by (Stat) 
  26] s(X) > X  because [27], by definition 
  27] s*(X) >= X  because [28], by (Select) 
  28] X >= X  by (Meta) 
  29] plus*(s(X), Y) >= X  because [30], by (Select) 
  30] s(X) >= X  because [27], by (Star) 
  31] plus*(s(X), Y) >= s(Y)  because plus > s and [32], by (Copy) 
  32] plus*(s(X), Y) >= Y  because [33], by (Select) 
  33] Y >= Y  by (Meta) 

  34] plus(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [35], by (Star) 
  35] plus*(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [36], [37] and [39], by (Stat) 
  36] plus(X, Y) > X  because [5], by definition 
  37] plus*(plus(X, Y), Z) >= X  because [38], by (Select) 
  38] plus(X, Y) >= X  because [5], by (Star) 
  39] plus*(plus(X, Y), Z) >= plus(Y, Z)  because [40], [42] and [43], by (Stat) 
  40] plus(X, Y) > Y  because [41], by definition 
  41] plus*(X, Y) >= Y  because [10], by (Select) 
  42] plus*(plus(X, Y), Z) >= Y  because [8], by (Select) 
  43] plus*(plus(X, Y), Z) >= Z  because [44], by (Select) 
  44] Z >= Z  by (Meta) 

  45] mult(_|_, X) >= _|_  by (Bot) 

  46] mult(s(X), Y) >= plus(mult(X, Y), Y)  because [47], by (Star) 
  47] mult*(s(X), Y) >= plus(mult(X, Y), Y)  because mult > plus, [48] and [52], by (Copy) 
  48] mult*(s(X), Y) >= mult(X, Y)  because mult in Mul, [49] and [51], by (Stat) 
  49] s(X) > X  because [50], by definition 
  50] s*(X) >= X  because [14], by (Select) 
  51] Y >= Y  by (Meta) 
  52] mult*(s(X), Y) >= Y  because [51], by (Select) 

  53] mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because [54], by (Star) 
  54] mult*(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because mult > plus, [55] and [58], by (Copy) 
  55] mult*(plus(X, Y), Z) >= mult(X, Z)  because mult in Mul, [56] and [57], by (Stat) 
  56] plus(X, Y) > X  because [19], by definition 
  57] Z >= Z  by (Meta) 
  58] mult*(plus(X, Y), Z) >= mult(Y, Z)  because mult in Mul, [59] and [57], by (Stat) 
  59] plus(X, Y) > Y  because [60], by definition 
  60] plus*(X, Y) >= Y  because [61], by (Select) 
  61] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_0, static, formative) by (P_4, R_0, static, formative), where P_4 consists of:

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

Thus, the original system is terminating if (P_4, R_0, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  plus#(plus(X, Y), Z) >? plus#(X, plus(Y, Z)) 
  plus#(plus(X, Y), Z) >? plus#(Y, Z) 
  mult#(s(X), Y) >? mult#(X, Y) 
  plus(z, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(z, X) >= z 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[mult#(x_1, x_2)]] = x_1 
  [[z]] = _|_ 

We choose Lex = {plus, plus#} and Mul = {mult, s}, and the following precedence: mult > plus# > plus > s

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  plus#(plus(X, Y), Z) >= plus#(X, plus(Y, Z)) 
  plus#(plus(X, Y), Z) > plus#(Y, Z) 
  s(X) >= X 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(_|_, X) >= _|_ 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

With these choices, we have:

  1] plus#(plus(X, Y), Z) >= plus#(X, plus(Y, Z))  because [2], by (Star) 
  2] plus#*(plus(X, Y), Z) >= plus#(X, plus(Y, Z))  because [3], [6] and [8], by (Stat) 
  3] plus(X, Y) > X  because [4], by definition 
  4] plus*(X, Y) >= X  because [5], by (Select) 
  5] X >= X  by (Meta) 
  6] plus#*(plus(X, Y), Z) >= X  because [7], by (Select) 
  7] plus(X, Y) >= X  because [4], by (Star) 
  8] plus#*(plus(X, Y), Z) >= plus(Y, Z)  because plus# > plus, [9] and [13], by (Copy) 
  9] plus#*(plus(X, Y), Z) >= Y  because [10], by (Select) 
  10] plus(X, Y) >= Y  because [11], by (Star) 
  11] plus*(X, Y) >= Y  because [12], by (Select) 
  12] Y >= Y  by (Meta) 
  13] plus#*(plus(X, Y), Z) >= Z  because [14], by (Select) 
  14] Z >= Z  by (Meta) 

  15] plus#(plus(X, Y), Z) > plus#(Y, Z)  because [16], by definition 
  16] plus#*(plus(X, Y), Z) >= plus#(Y, Z)  because [17], [9] and [13], by (Stat) 
  17] plus(X, Y) > Y  because [18], by definition 
  18] plus*(X, Y) >= Y  because [12], by (Select) 

  19] s(X) >= X  because [20], by (Star) 
  20] s*(X) >= X  because [21], by (Select) 
  21] X >= X  by (Meta) 

  22] plus(_|_, X) >= X  because [23], by (Star) 
  23] plus*(_|_, X) >= X  because [24], by (Select) 
  24] X >= X  by (Meta) 

  25] plus(s(X), Y) >= plus(X, s(Y))  because [26], by (Star) 
  26] plus*(s(X), Y) >= plus(X, s(Y))  because [27], [30] and [32], by (Stat) 
  27] s(X) > X  because [28], by definition 
  28] s*(X) >= X  because [29], by (Select) 
  29] X >= X  by (Meta) 
  30] plus*(s(X), Y) >= X  because [31], by (Select) 
  31] s(X) >= X  because [28], by (Star) 
  32] plus*(s(X), Y) >= s(Y)  because plus > s and [33], by (Copy) 
  33] plus*(s(X), Y) >= Y  because [34], by (Select) 
  34] Y >= Y  by (Meta) 

  35] plus(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [36], by (Star) 
  36] plus*(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [3], [37] and [38], by (Stat) 
  37] plus*(plus(X, Y), Z) >= X  because [7], by (Select) 
  38] plus*(plus(X, Y), Z) >= plus(Y, Z)  because [17], [39] and [40], by (Stat) 
  39] plus*(plus(X, Y), Z) >= Y  because [10], by (Select) 
  40] plus*(plus(X, Y), Z) >= Z  because [14], by (Select) 

  41] mult(_|_, X) >= _|_  by (Bot) 

  42] mult(s(X), Y) >= plus(mult(X, Y), Y)  because [43], by (Star) 
  43] mult*(s(X), Y) >= plus(mult(X, Y), Y)  because mult > plus, [44] and [48], by (Copy) 
  44] mult*(s(X), Y) >= mult(X, Y)  because mult in Mul, [45] and [47], by (Stat) 
  45] s(X) > X  because [46], by definition 
  46] s*(X) >= X  because [21], by (Select) 
  47] Y >= Y  by (Meta) 
  48] mult*(s(X), Y) >= Y  because [47], by (Select) 

  49] mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because [50], by (Star) 
  50] mult*(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because mult > plus, [51] and [56], by (Copy) 
  51] mult*(plus(X, Y), Z) >= mult(X, Z)  because mult in Mul, [52] and [55], by (Stat) 
  52] plus(X, Y) > X  because [53], by definition 
  53] plus*(X, Y) >= X  because [54], by (Select) 
  54] X >= X  by (Meta) 
  55] Z >= Z  by (Meta) 
  56] mult*(plus(X, Y), Z) >= mult(Y, Z)  because mult in Mul, [57] and [55], by (Stat) 
  57] plus(X, Y) > Y  because [58], by definition 
  58] plus*(X, Y) >= Y  because [59], by (Select) 
  59] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_0, static, formative) by (P_5, R_0, static, formative), where P_5 consists of:

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

Thus, the original system is terminating if (P_5, R_0, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  plus#(plus(X, Y), Z) >? plus#(X, plus(Y, Z)) 
  mult#(s(X), Y) >? mult#(X, Y) 
  plus(z, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(z, X) >= z 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[plus#(x_1, x_2)]] = x_1 
  [[z]] = _|_ 

We choose Lex = {plus} and Mul = {mult, mult#, s}, and the following precedence: mult# > mult > plus > s

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  plus(X, Y) > X 
  mult#(s(X), Y) >= mult#(X, Y) 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(_|_, X) >= _|_ 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

With these choices, we have:

  1] plus(X, Y) > X  because [2], by definition 
  2] plus*(X, Y) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] mult#(s(X), Y) >= mult#(X, Y)  because mult# in Mul, [5] and [8], by (Fun) 
  5] s(X) >= X  because [6], by (Star) 
  6] s*(X) >= X  because [7], by (Select) 
  7] X >= X  by (Meta) 
  8] Y >= Y  by (Meta) 

  9] plus(_|_, X) >= X  because [10], by (Star) 
  10] plus*(_|_, X) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 

  12] plus(s(X), Y) >= plus(X, s(Y))  because [13], by (Star) 
  13] plus*(s(X), Y) >= plus(X, s(Y))  because [14], [17] and [19], by (Stat) 
  14] s(X) > X  because [15], by definition 
  15] s*(X) >= X  because [16], by (Select) 
  16] X >= X  by (Meta) 
  17] plus*(s(X), Y) >= X  because [18], by (Select) 
  18] s(X) >= X  because [15], by (Star) 
  19] plus*(s(X), Y) >= s(Y)  because plus > s and [20], by (Copy) 
  20] plus*(s(X), Y) >= Y  because [21], by (Select) 
  21] Y >= Y  by (Meta) 

  22] plus(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [23], by (Star) 
  23] plus*(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [24], [25] and [27], by (Stat) 
  24] plus(X, Y) > X  because [2], by definition 
  25] plus*(plus(X, Y), Z) >= X  because [26], by (Select) 
  26] plus(X, Y) >= X  because [2], by (Star) 
  27] plus*(plus(X, Y), Z) >= plus(Y, Z)  because [28], [31] and [33], by (Stat) 
  28] plus(X, Y) > Y  because [29], by definition 
  29] plus*(X, Y) >= Y  because [30], by (Select) 
  30] Y >= Y  by (Meta) 
  31] plus*(plus(X, Y), Z) >= Y  because [32], by (Select) 
  32] plus(X, Y) >= Y  because [29], by (Star) 
  33] plus*(plus(X, Y), Z) >= Z  because [34], by (Select) 
  34] Z >= Z  by (Meta) 

  35] mult(_|_, X) >= _|_  by (Bot) 

  36] mult(s(X), Y) >= plus(mult(X, Y), Y)  because [37], by (Star) 
  37] mult*(s(X), Y) >= plus(mult(X, Y), Y)  because mult > plus, [38] and [41], by (Copy) 
  38] mult*(s(X), Y) >= mult(X, Y)  because mult in Mul, [39] and [8], by (Stat) 
  39] s(X) > X  because [40], by definition 
  40] s*(X) >= X  because [7], by (Select) 
  41] mult*(s(X), Y) >= Y  because [8], by (Select) 

  42] mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because [43], by (Star) 
  43] mult*(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because mult > plus, [44] and [49], by (Copy) 
  44] mult*(plus(X, Y), Z) >= mult(X, Z)  because mult in Mul, [45] and [48], by (Stat) 
  45] plus(X, Y) > X  because [46], by definition 
  46] plus*(X, Y) >= X  because [47], by (Select) 
  47] X >= X  by (Meta) 
  48] Z >= Z  by (Meta) 
  49] mult*(plus(X, Y), Z) >= mult(Y, Z)  because mult in Mul, [50] and [48], by (Stat) 
  50] plus(X, Y) > Y  because [51], by definition 
  51] plus*(X, Y) >= Y  because [52], by (Select) 
  52] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_0, static, formative) by (P_6, R_0, static, formative), where P_6 consists of:

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

Thus, the original system is terminating if (P_6, R_0, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  mult#(s(X), Y) >? mult#(X, Y) 
  plus(z, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(z, X) >= z 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[mult#(x_1, x_2)]] = x_1 
  [[z]] = _|_ 

We choose Lex = {plus} and Mul = {mult, s}, and the following precedence: mult > plus > s

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  s(X) > X 
  plus(_|_, X) >= X 
  plus(s(X), Y) >= plus(X, s(Y)) 
  plus(plus(X, Y), Z) >= plus(X, plus(Y, Z)) 
  mult(_|_, X) >= _|_ 
  mult(s(X), Y) >= plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z)) 

With these choices, we have:

  1] s(X) > X  because [2], by definition 
  2] s*(X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] plus(_|_, X) >= X  because [5], by (Star) 
  5] plus*(_|_, X) >= X  because [6], by (Select) 
  6] X >= X  by (Meta) 

  7] plus(s(X), Y) >= plus(X, s(Y))  because [8], by (Star) 
  8] plus*(s(X), Y) >= plus(X, s(Y))  because [9], [12] and [14], by (Stat) 
  9] s(X) > X  because [10], by definition 
  10] s*(X) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 
  12] plus*(s(X), Y) >= X  because [13], by (Select) 
  13] s(X) >= X  because [10], by (Star) 
  14] plus*(s(X), Y) >= s(Y)  because plus > s and [15], by (Copy) 
  15] plus*(s(X), Y) >= Y  because [16], by (Select) 
  16] Y >= Y  by (Meta) 

  17] plus(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [18], by (Star) 
  18] plus*(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [19], [22] and [24], by (Stat) 
  19] plus(X, Y) > X  because [20], by definition 
  20] plus*(X, Y) >= X  because [21], by (Select) 
  21] X >= X  by (Meta) 
  22] plus*(plus(X, Y), Z) >= X  because [23], by (Select) 
  23] plus(X, Y) >= X  because [20], by (Star) 
  24] plus*(plus(X, Y), Z) >= plus(Y, Z)  because [25], [28] and [30], by (Stat) 
  25] plus(X, Y) > Y  because [26], by definition 
  26] plus*(X, Y) >= Y  because [27], by (Select) 
  27] Y >= Y  by (Meta) 
  28] plus*(plus(X, Y), Z) >= Y  because [29], by (Select) 
  29] plus(X, Y) >= Y  because [26], by (Star) 
  30] plus*(plus(X, Y), Z) >= Z  because [31], by (Select) 
  31] Z >= Z  by (Meta) 

  32] mult(_|_, X) >= _|_  by (Bot) 

  33] mult(s(X), Y) >= plus(mult(X, Y), Y)  because [34], by (Star) 
  34] mult*(s(X), Y) >= plus(mult(X, Y), Y)  because mult > plus, [35] and [38], by (Copy) 
  35] mult*(s(X), Y) >= mult(X, Y)  because mult in Mul, [36] and [37], by (Stat) 
  36] s(X) > X  because [2], by definition 
  37] Y >= Y  by (Meta) 
  38] mult*(s(X), Y) >= Y  because [37], by (Select) 

  39] mult(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because [40], by (Star) 
  40] mult*(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because mult > plus, [41] and [46], by (Copy) 
  41] mult*(plus(X, Y), Z) >= mult(X, Z)  because mult in Mul, [42] and [45], by (Stat) 
  42] plus(X, Y) > X  because [43], by definition 
  43] plus*(X, Y) >= X  because [44], by (Select) 
  44] X >= X  by (Meta) 
  45] Z >= Z  by (Meta) 
  46] mult*(plus(X, Y), Z) >= mult(Y, Z)  because mult in Mul, [47] and [45], by (Stat) 
  47] plus(X, Y) > Y  because [48], by definition 
  48] plus*(X, Y) >= Y  because [49], by (Select) 
  49] Y >= Y  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_6, R_0) by ({}, R_0).  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 +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
[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.