We consider the system AotoYamada_05__014. Alphabet: 0 : [] --> b cons : [b * a] --> a double : [] --> a -> a inc : [] --> a -> a map : [b -> b] --> a -> a nil : [] --> a plus : [b] --> b -> b s : [b] --> b times : [b] --> b -> b Rules: plus(0) x => x plus(s(x)) y => s(plus(x) y) times(0) x => 0 times(s(x)) y => plus(times(x) y) y map(f) nil => nil map(f) cons(x, y) => cons(f x, map(f) y) inc => map(plus(s(0))) double => map(times(s(s(0)))) 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]). In order to do so, we start by eta-expanding the system, which gives: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(X), Y) => plus(times(X, Y), Y) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) inc(X) => map(/\x.plus(s(0), x), X) double(X) => map(/\x.times(s(s(0)), x), X) 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, Y) 1] times#(s(X), Y) =#> plus#(times(X, Y), Y) 2] times#(s(X), Y) =#> times#(X, Y) 3] map#(F, cons(X, Y)) =#> map#(F, Y) 4] inc#(X) =#> map#(/\x.plus(s(0), x), X) 5] inc#(X) =#> plus#(s(0), Y) 6] double#(X) =#> map#(/\x.times(s(s(0)), x), X) 7] double#(X) =#> times#(s(s(0)), Y) Rules R_0: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(X), Y) => plus(times(X, Y), Y) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) inc(X) => map(/\x.plus(s(0), x), X) double(X) => map(/\x.times(s(s(0)), x), X) 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, Y) times#(s(X), Y) >? plus#(times(X, Y), Y) times#(s(X), Y) >? times#(X, Y) map#(F, cons(X, Y)) >? map#(F, Y) inc#(X) >? map#(/\x.plus(s(0), x), X) inc#(X) >? plus#(s(0), Y) double#(X) >? map#(/\x.times(s(s(0)), x), X) double#(X) >? times#(s(s(0)), Y) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) times(0, X) >= 0 times(s(X), Y) >= plus(times(X, Y), Y) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) inc(X) >= map(/\x.plus(s(0), x), X) double(X) >= map(/\x.times(s(s(0)), x), X) We apply [Kop12, Thm. 6.75] and use the following argument functions: pi( double(X) ) = #argfun-double#(map(/\x.times(s(s(0)), x), X)) pi( inc(X) ) = #argfun-inc#(map(/\x.plus(s(0), x), X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: #argfun-double# = \y0.3 + y0 #argfun-inc# = \y0.3 + y0 0 = 0 cons = \y0y1.2 double = \y0.0 double# = \y0.3 + 2y0 inc = \y0.0 inc# = \y0.3 + 2y0 map = \G0y1.2 + 2y1G0(y1) + 2G0(0) map# = \G0y1.0 nil = 0 plus = \y0y1.2 + y1 plus# = \y0y1.0 s = \y0.0 times = \y0y1.2 + y1 times# = \y0y1.0 Using this interpretation, the requirements translate to: [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] [[times#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(times(_x0, _x1), _x1)]] [[times#(s(_x0), _x1)]] = 0 >= 0 = [[times#(_x0, _x1)]] [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] [[inc#(_x0)]] = 3 + 2x0 > 0 = [[map#(/\x.plus(s(0), x), _x0)]] [[inc#(_x0)]] = 3 + 2x0 > 0 = [[plus#(s(0), _x1)]] [[double#(_x0)]] = 3 + 2x0 > 0 = [[map#(/\x.times(s(s(0)), x), _x0)]] [[double#(_x0)]] = 3 + 2x0 > 0 = [[times#(s(s(0)), _x1)]] [[plus(0, _x0)]] = 2 + x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = 2 + x1 >= 0 = [[s(plus(_x0, _x1))]] [[times(0, _x0)]] = 2 + x0 >= 0 = [[0]] [[times(s(_x0), _x1)]] = 2 + x1 >= 2 + x1 = [[plus(times(_x0, _x1), _x1)]] [[map(_F0, nil)]] = 2 + 2F0(0) >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 2 + 2F0(0) + 4F0(2) >= 2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[#argfun-inc#(map(/\x.plus(s(0), x), _x0))]] = 9 + 2x0x0 + 4x0 >= 6 + 2x0x0 + 4x0 = [[map(/\x.plus(s(0), x), _x0)]] [[#argfun-double#(map(/\x.times(s(s(0)), x), _x0))]] = 9 + 2x0x0 + 4x0 >= 6 + 2x0x0 + 4x0 = [[map(/\x.times(s(s(0)), x), _x0)]] 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, Y) times#(s(X), Y) =#> plus#(times(X, Y), Y) times#(s(X), Y) =#> times#(X, Y) map#(F, cons(X, Y)) =#> map#(F, Y) 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, Y) times#(s(X), Y) >? plus#(times(X, Y), Y) times#(s(X), Y) >? times#(X, Y) map#(F, cons(X, Y)) >? map#(F, Y) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) times(0, X) >= 0 times(s(X), Y) >= plus(times(X, Y), Y) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) inc(X) >= map(/\x.plus(s(0), x), X) double(X) >= map(/\x.times(s(s(0)), x), X) We apply [Kop12, Thm. 6.75] and use the following argument functions: pi( double(X) ) = #argfun-double#(map(/\x.times(s(s(0)), x), X)) pi( inc(X) ) = #argfun-inc#(map(/\x.plus(s(0), x), X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: #argfun-double# = \y0.3 + y0 #argfun-inc# = \y0.3 + y0 0 = 0 cons = \y0y1.3 + y0 + y1 double = \y0.0 inc = \y0.0 map = \G0y1.y1 + 2G0(0) + 3y1G0(y1) map# = \G0y1.0 nil = 0 plus = \y0y1.y1 plus# = \y0y1.0 s = \y0.0 times = \y0y1.y1 times# = \y0y1.3 + 2y1 Using this interpretation, the requirements translate to: [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] [[times#(s(_x0), _x1)]] = 3 + 2x1 > 0 = [[plus#(times(_x0, _x1), _x1)]] [[times#(s(_x0), _x1)]] = 3 + 2x1 >= 3 + 2x1 = [[times#(_x0, _x1)]] [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = x1 >= 0 = [[s(plus(_x0, _x1))]] [[times(0, _x0)]] = x0 >= 0 = [[0]] [[times(s(_x0), _x1)]] = x1 >= x1 = [[plus(times(_x0, _x1), _x1)]] [[map(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + 2F0(0) + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 9F0(3 + x1 + x2) >= 3 + x2 + F0(x1) + 2F0(0) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[#argfun-inc#(map(/\x.plus(s(0), x), _x0))]] = 3 + x0 + 3x0x0 >= x0 + 3x0x0 = [[map(/\x.plus(s(0), x), _x0)]] [[#argfun-double#(map(/\x.times(s(s(0)), x), _x0))]] = 3 + x0 + 3x0x0 >= x0 + 3x0x0 = [[map(/\x.times(s(s(0)), x), _x0)]] 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#(s(X), Y) =#> plus#(X, Y) times#(s(X), Y) =#> times#(X, Y) map#(F, cons(X, Y)) =#> map#(F, Y) 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#(s(X), Y) >? plus#(X, Y) times#(s(X), Y) >? times#(X, Y) map#(F, cons(X, Y)) >? map#(F, Y) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) times(0, X) >= 0 times(s(X), Y) >= plus(times(X, Y), Y) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) inc(X) >= map(/\x.plus(s(0), x), X) double(X) >= map(/\x.times(s(s(0)), x), X) We apply [Kop12, Thm. 6.75] and use the following argument functions: pi( double(X) ) = #argfun-double#(map(/\x.times(s(s(0)), x), X)) pi( inc(X) ) = #argfun-inc#(map(/\x.plus(s(0), x), X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: #argfun-double# = \y0.3 + y0 #argfun-inc# = \y0.3 + y0 0 = 0 cons = \y0y1.1 + y1 double = \y0.0 inc = \y0.0 map = \G0y1.y1 map# = \G0y1.y1 nil = 0 plus = \y0y1.y1 plus# = \y0y1.0 s = \y0.0 times = \y0y1.y1 times# = \y0y1.0 Using this interpretation, the requirements translate to: [[plus#(s(_x0), _x1)]] = 0 >= 0 = [[plus#(_x0, _x1)]] [[times#(s(_x0), _x1)]] = 0 >= 0 = [[times#(_x0, _x1)]] [[map#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[map#(_F0, _x2)]] [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = x1 >= 0 = [[s(plus(_x0, _x1))]] [[times(0, _x0)]] = x0 >= 0 = [[0]] [[times(s(_x0), _x1)]] = x1 >= x1 = [[plus(times(_x0, _x1), _x1)]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[#argfun-inc#(map(/\x.plus(s(0), x), _x0))]] = 3 + x0 >= x0 = [[map(/\x.plus(s(0), x), _x0)]] [[#argfun-double#(map(/\x.times(s(s(0)), x), _x0))]] = 3 + x0 >= x0 = [[map(/\x.times(s(s(0)), x), _x0)]] 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#(s(X), Y) =#> plus#(X, Y) times#(s(X), Y) =#> times#(X, Y) 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#(s(X), Y) >? plus#(X, Y) times#(s(X), Y) >? times#(X, Y) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) times(0, X) >= 0 times(s(X), Y) >= plus(times(X, Y), Y) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) inc(X) >= map(/\x.plus(s(0), x), X) double(X) >= map(/\x.times(s(s(0)), x), X) We apply [Kop12, Thm. 6.75] and use the following argument functions: pi( double(X) ) = #argfun-double#(map(/\x.times(s(s(0)), x), X)) pi( inc(X) ) = #argfun-inc#(map(/\x.plus(s(0), x), X)) 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: [[#argfun-inc#(x_1)]] = #argfun-inc# [[0]] = _|_ [[cons(x_1, x_2)]] = x_2 [[double(x_1)]] = double [[inc(x_1)]] = inc [[map(x_1, x_2)]] = _|_ [[nil]] = _|_ We choose Lex = {} and Mul = {#argfun-double#, #argfun-inc#, @_{o -> o}, double, inc, plus, plus#, s, times, times#}, and the following precedence: #argfun-inc# > @_{o -> o} > #argfun-double# > double > inc > plus# > times > plus > s > times# Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: plus#(s(X), Y) > plus#(X, Y) times#(s(X), Y) >= times#(X, Y) plus(_|_, X) >= X plus(s(X), Y) >= s(plus(X, Y)) times(_|_, X) >= _|_ times(s(X), Y) >= plus(times(X, Y), Y) _|_ >= _|_ _|_ >= _|_ #argfun-inc# >= _|_ #argfun-double#(_|_) >= _|_ With these choices, we have: 1] plus#(s(X), Y) > plus#(X, Y) because [2], by definition 2] plus#*(s(X), Y) >= plus#(X, Y) because plus# in Mul, [3] and [6], by (Stat) 3] s(X) > X because [4], by definition 4] s*(X) >= X because [5], by (Select) 5] X >= X by (Meta) 6] Y >= Y by (Meta) 7] times#(s(X), Y) >= times#(X, Y) because times# in Mul, [8] and [11], by (Fun) 8] s(X) >= X because [9], by (Star) 9] s*(X) >= X because [10], by (Select) 10] X >= X by (Meta) 11] Y >= Y by (Meta) 12] plus(_|_, X) >= X because [13], by (Star) 13] plus*(_|_, X) >= X because [14], by (Select) 14] X >= X by (Meta) 15] plus(s(X), Y) >= s(plus(X, Y)) because [16], by (Star) 16] plus*(s(X), Y) >= s(plus(X, Y)) because plus > s and [17], by (Copy) 17] plus*(s(X), Y) >= plus(X, Y) because plus in Mul, [3] and [6], by (Stat) 18] times(_|_, X) >= _|_ by (Bot) 19] times(s(X), Y) >= plus(times(X, Y), Y) because [20], by (Star) 20] times*(s(X), Y) >= plus(times(X, Y), Y) because times > plus, [21] and [24], by (Copy) 21] times*(s(X), Y) >= times(X, Y) because times in Mul, [22] and [11], by (Stat) 22] s(X) > X because [23], by definition 23] s*(X) >= X because [10], by (Select) 24] times*(s(X), Y) >= Y because [11], by (Select) 25] _|_ >= _|_ by (Bot) 26] _|_ >= _|_ by (Bot) 27] #argfun-inc# >= _|_ by (Bot) 28] #argfun-double#(_|_) >= _|_ by (Bot) 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: times#(s(X), Y) =#> times#(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: times#(s(X), Y) >? times#(X, Y) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) times(0, X) >= 0 times(s(X), Y) >= plus(times(X, Y), Y) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) inc(X) >= map(/\x.plus(s(0), x), X) double(X) >= map(/\x.times(s(s(0)), x), X) We apply [Kop12, Thm. 6.75] and use the following argument functions: pi( double(X) ) = #argfun-double#(map(/\x.times(s(s(0)), x), X)) pi( inc(X) ) = #argfun-inc#(map(/\x.plus(s(0), x), X)) 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: [[0]] = _|_ [[double(x_1)]] = x_1 [[inc(x_1)]] = x_1 [[nil]] = _|_ [[times#(x_1, x_2)]] = times#(x_1) We choose Lex = {} and Mul = {#argfun-double#, #argfun-inc#, @_{o -> o}, cons, map, plus, s, times, times#}, and the following precedence: #argfun-double# > #argfun-inc# > times > map > @_{o -> o} > cons > plus > s = times# Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: times#(s(X)) > times#(X) plus(_|_, X) >= X plus(s(X), Y) >= s(plus(X, Y)) times(_|_, X) >= _|_ times(s(X), Y) >= plus(times(X, Y), Y) map(F, _|_) >= _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) #argfun-inc#(map(/\x.plus(s(_|_), x), X)) >= map(/\x.plus(s(_|_), x), X) #argfun-double#(map(/\x.times(s(s(_|_)), x), X)) >= map(/\x.times(s(s(_|_)), x), X) With these choices, we have: 1] times#(s(X)) > times#(X) because [2], by definition 2] times#*(s(X)) >= times#(X) because [3], by (Select) 3] s(X) >= times#(X) because s = times#, s in Mul and [4], by (Fun) 4] X >= X by (Meta) 5] plus(_|_, X) >= X because [6], by (Star) 6] plus*(_|_, X) >= X because [7], by (Select) 7] X >= X by (Meta) 8] plus(s(X), Y) >= s(plus(X, Y)) because [9], by (Star) 9] plus*(s(X), Y) >= s(plus(X, Y)) because plus > s and [10], by (Copy) 10] plus*(s(X), Y) >= plus(X, Y) because plus 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] times(_|_, X) >= _|_ by (Bot) 16] times(s(X), Y) >= plus(times(X, Y), Y) because [17], by (Star) 17] times*(s(X), Y) >= plus(times(X, Y), Y) because times > plus, [18] and [22], by (Copy) 18] times*(s(X), Y) >= times(X, Y) because times in Mul, [19] and [21], by (Stat) 19] s(X) > X because [20], by definition 20] s*(X) >= X because [4], by (Select) 21] Y >= Y by (Meta) 22] times*(s(X), Y) >= Y because [21], by (Select) 23] map(F, _|_) >= _|_ by (Bot) 24] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [25], by (Star) 25] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [26] and [33], by (Copy) 26] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [27] and [29], by (Copy) 27] map*(F, cons(X, Y)) >= F because [28], by (Select) 28] F >= F by (Meta) 29] map*(F, cons(X, Y)) >= X because [30], by (Select) 30] cons(X, Y) >= X because [31], by (Star) 31] cons*(X, Y) >= X because [32], by (Select) 32] X >= X by (Meta) 33] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [34] and [35], by (Stat) 34] F >= F by (Meta) 35] cons(X, Y) > Y because [36], by definition 36] cons*(X, Y) >= Y because [37], by (Select) 37] Y >= Y by (Meta) 38] #argfun-inc#(map(/\x.plus(s(_|_), x), X)) >= map(/\x.plus(s(_|_), x), X) because [39], by (Star) 39] #argfun-inc#*(map(/\x.plus(s(_|_), x), X)) >= map(/\x.plus(s(_|_), x), X) because #argfun-inc# > map, [40] and [46], by (Copy) 40] #argfun-inc#*(map(/\x.plus(s(_|_), x), X)) >= /\x.plus(s(_|_), x) because [41], by (F-Abs) 41] #argfun-inc#*(map(/\x.plus(s(_|_), x), X), y) >= plus(s(_|_), y) because #argfun-inc# > plus, [42] and [44], by (Copy) 42] #argfun-inc#*(map(/\x.plus(s(_|_), x), X), y) >= s(_|_) because #argfun-inc# > s and [43], by (Copy) 43] #argfun-inc#*(map(/\x.plus(s(_|_), x), X), y) >= _|_ by (Bot) 44] #argfun-inc#*(map(/\x.plus(s(_|_), x), X), y) >= y because [45], by (Select) 45] y >= y by (Var) 46] #argfun-inc#*(map(/\x.plus(s(_|_), x), X)) >= X because [47], by (Select) 47] map(/\x.plus(s(_|_), x), X) >= X because [48], by (Star) 48] map*(/\x.plus(s(_|_), x), X) >= X because [49], by (Select) 49] X >= X by (Meta) 50] #argfun-double#(map(/\x.times(s(s(_|_)), x), X)) >= map(/\x.times(s(s(_|_)), x), X) because [51], by (Star) 51] #argfun-double#*(map(/\x.times(s(s(_|_)), x), X)) >= map(/\x.times(s(s(_|_)), x), X) because #argfun-double# > map, [52] and [59], by (Copy) 52] #argfun-double#*(map(/\x.times(s(s(_|_)), x), X)) >= /\x.times(s(s(_|_)), x) because [53], by (F-Abs) 53] #argfun-double#*(map(/\x.times(s(s(_|_)), x), X), y) >= times(s(s(_|_)), y) because #argfun-double# > times, [54] and [57], by (Copy) 54] #argfun-double#*(map(/\x.times(s(s(_|_)), x), X), y) >= s(s(_|_)) because #argfun-double# > s and [55], by (Copy) 55] #argfun-double#*(map(/\x.times(s(s(_|_)), x), X), y) >= s(_|_) because #argfun-double# > s and [56], by (Copy) 56] #argfun-double#*(map(/\x.times(s(s(_|_)), x), X), y) >= _|_ by (Bot) 57] #argfun-double#*(map(/\x.times(s(s(_|_)), x), X), y) >= y because [58], by (Select) 58] y >= y by (Var) 59] #argfun-double#*(map(/\x.times(s(s(_|_)), x), X)) >= X because [60], by (Select) 60] map(/\x.times(s(s(_|_)), x), X) >= X because [61], by (Star) 61] map*(/\x.times(s(s(_|_)), x), X) >= X because [62], by (Select) 62] X >= X by (Meta) 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. 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.