We consider the system AotoYamada_05__021. Alphabet: 0 : [] --> a cons : [a * b] --> b double : [b] --> b inc : [b] --> b map : [a -> a * b] --> b nil : [] --> b plus : [a] --> a -> a s : [a] --> a times : [a] --> a -> a 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 inc(x) => map(plus(s(0)), x) double(x) => map(times(s(s(0))), x) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 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 inc(X) >? map(plus(s(0)), X) double(X) >? map(times(s(s(0))), X) map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[nil]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, cons, double, inc, map, plus, s, times}, and the following precedence: double > inc > map > cons > times > plus > @_{o -> o} > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: @_{o -> o}(plus(_|_), X) > X @_{o -> o}(plus(s(X)), Y) > s(@_{o -> o}(plus(X), Y)) @_{o -> o}(times(_|_), X) >= _|_ @_{o -> o}(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) inc(X) >= map(plus(s(_|_)), X) double(X) >= map(times(s(s(_|_))), X) map(F, _|_) >= _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) With these choices, we have: 1] @_{o -> o}(plus(_|_), X) > X because [2], by definition 2] @_{o -> o}*(plus(_|_), X) >= X because [3], by (Select) 3] X >= X by (Meta) 4] @_{o -> o}(plus(s(X)), Y) > s(@_{o -> o}(plus(X), Y)) because [5], by definition 5] @_{o -> o}*(plus(s(X)), Y) >= s(@_{o -> o}(plus(X), Y)) because [6], by (Select) 6] plus(s(X)) @_{o -> o}*(plus(s(X)), Y) >= s(@_{o -> o}(plus(X), Y)) because [7] 7] plus*(s(X), @_{o -> o}*(plus(s(X)), Y)) >= s(@_{o -> o}(plus(X), Y)) because plus > s and [8], by (Copy) 8] plus*(s(X), @_{o -> o}*(plus(s(X)), Y)) >= @_{o -> o}(plus(X), Y) because plus > @_{o -> o}, [9] and [13], by (Copy) 9] plus*(s(X), @_{o -> o}*(plus(s(X)), Y)) >= plus(X) because plus in Mul and [10], by (Stat) 10] s(X) > X because [11], by definition 11] s*(X) >= X because [12], by (Select) 12] X >= X by (Meta) 13] plus*(s(X), @_{o -> o}*(plus(s(X)), Y)) >= Y because [14], by (Select) 14] @_{o -> o}*(plus(s(X)), Y) >= Y because [15], by (Select) 15] Y >= Y by (Meta) 16] @_{o -> o}(times(_|_), X) >= _|_ by (Bot) 17] @_{o -> o}(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) because [18], by (Star) 18] @_{o -> o}*(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) because [19], by (Select) 19] times(s(X)) @_{o -> o}*(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) because [20] 20] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) because times > @_{o -> o}, [21] and [27], by (Copy) 21] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= plus(@_{o -> o}(times(X), Y)) because times > plus and [22], by (Copy) 22] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= @_{o -> o}(times(X), Y) because times > @_{o -> o}, [23] and [27], by (Copy) 23] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= times(X) because times in Mul and [24], by (Stat) 24] s(X) > X because [25], by definition 25] s*(X) >= X because [26], by (Select) 26] X >= X by (Meta) 27] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= Y because [28], by (Select) 28] @_{o -> o}*(times(s(X)), Y) >= Y because [29], by (Select) 29] Y >= Y by (Meta) 30] inc(X) >= map(plus(s(_|_)), X) because [31], by (Star) 31] inc*(X) >= map(plus(s(_|_)), X) because inc > map, [32] and [35], by (Copy) 32] inc*(X) >= plus(s(_|_)) because inc > plus and [33], by (Copy) 33] inc*(X) >= s(_|_) because inc > s and [34], by (Copy) 34] inc*(X) >= _|_ by (Bot) 35] inc*(X) >= X because [36], by (Select) 36] X >= X by (Meta) 37] double(X) >= map(times(s(s(_|_))), X) because [38], by (Star) 38] double*(X) >= map(times(s(s(_|_))), X) because double > map, [39] and [43], by (Copy) 39] double*(X) >= times(s(s(_|_))) because double > times and [40], by (Copy) 40] double*(X) >= s(s(_|_)) because double > s and [41], by (Copy) 41] double*(X) >= s(_|_) because double > s and [42], by (Copy) 42] double*(X) >= _|_ by (Bot) 43] double*(X) >= X because [44], by (Select) 44] X >= X by (Meta) 45] map(F, _|_) >= _|_ by (Bot) 46] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [47], by (Star) 47] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [48] and [55], by (Copy) 48] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [49] and [51], by (Copy) 49] map*(F, cons(X, Y)) >= F because [50], by (Select) 50] F >= F by (Meta) 51] map*(F, cons(X, Y)) >= X because [52], by (Select) 52] cons(X, Y) >= X because [53], by (Star) 53] cons*(X, Y) >= X because [54], by (Select) 54] X >= X by (Meta) 55] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [56] and [57], by (Stat) 56] F >= F by (Meta) 57] cons(X, Y) > Y because [58], by definition 58] cons*(X, Y) >= Y because [59], by (Select) 59] Y >= Y by (Meta) We can thus remove the following rules: plus(0) X => X plus(s(X)) Y => s(plus(X) Y) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): times(0) X >? 0 times(s(X)) Y >? plus(times(X) Y) Y inc(X) >? map(plus(s(0)), X) double(X) >? map(times(s(s(0))), X) map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[nil]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, cons, double, inc, map, plus, s, times}, and the following precedence: double > inc > s > times > @_{o -> o} = map > cons > plus Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: @_{o -> o}(times(_|_), X) >= _|_ @_{o -> o}(times(s(X)), Y) > @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) inc(X) >= map(plus(s(_|_)), X) double(X) >= map(times(s(s(_|_))), X) map(F, _|_) >= _|_ map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) With these choices, we have: 1] @_{o -> o}(times(_|_), X) >= _|_ by (Bot) 2] @_{o -> o}(times(s(X)), Y) > @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) because [3], by definition 3] @_{o -> o}*(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) because [4], by (Select) 4] times(s(X)) @_{o -> o}*(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) because [5] 5] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) because times > @_{o -> o}, [6] and [12], by (Copy) 6] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= plus(@_{o -> o}(times(X), Y)) because times > plus and [7], by (Copy) 7] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= @_{o -> o}(times(X), Y) because times > @_{o -> o}, [8] and [12], by (Copy) 8] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= times(X) because times in Mul and [9], by (Stat) 9] s(X) > X because [10], by definition 10] s*(X) >= X because [11], by (Select) 11] X >= X by (Meta) 12] times*(s(X), @_{o -> o}*(times(s(X)), Y)) >= Y because [13], by (Select) 13] @_{o -> o}*(times(s(X)), Y) >= Y because [14], by (Select) 14] Y >= Y by (Meta) 15] inc(X) >= map(plus(s(_|_)), X) because [16], by (Star) 16] inc*(X) >= map(plus(s(_|_)), X) because inc > map, [17] and [20], by (Copy) 17] inc*(X) >= plus(s(_|_)) because inc > plus and [18], by (Copy) 18] inc*(X) >= s(_|_) because inc > s and [19], by (Copy) 19] inc*(X) >= _|_ by (Bot) 20] inc*(X) >= X because [21], by (Select) 21] X >= X by (Meta) 22] double(X) >= map(times(s(s(_|_))), X) because [23], by (Star) 23] double*(X) >= map(times(s(s(_|_))), X) because double > map, [24] and [28], by (Copy) 24] double*(X) >= times(s(s(_|_))) because double > times and [25], by (Copy) 25] double*(X) >= s(s(_|_)) because double > s and [26], by (Copy) 26] double*(X) >= s(_|_) because double > s and [27], by (Copy) 27] double*(X) >= _|_ by (Bot) 28] double*(X) >= X because [29], by (Select) 29] X >= X by (Meta) 30] map(F, _|_) >= _|_ by (Bot) 31] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) because [32], by definition 32] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [33] and [38], by (Copy) 33] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map = @_{o -> o}, map in Mul, [34] and [35], by (Stat) 34] F >= F by (Meta) 35] cons(X, Y) > X because [36], by definition 36] cons*(X, Y) >= X because [37], by (Select) 37] X >= X by (Meta) 38] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [34] and [39], by (Stat) 39] cons(X, Y) > Y because [40], by definition 40] cons*(X, Y) >= Y because [41], by (Select) 41] Y >= Y by (Meta) We can thus remove the following rules: times(s(X)) Y => plus(times(X) Y) Y map(F, cons(X, Y)) => cons(F X, map(F, Y)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): times(0) X >? 0 inc(X) >? map(plus(s(0)), X) double(X) >? map(times(s(s(0))), X) map(F, nil) >? nil We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 double = \y0.3 + 3y0 inc = \y0.3 + 3y0 map = \G0y1.3y1 + 2G0(y1) nil = 2 plus = \y0y1.y0 s = \y0.y0 times = \y0y1.2y0 Using this interpretation, the requirements translate to: [[times(0) _x0]] = x0 >= 0 = [[0]] [[inc(_x0)]] = 3 + 3x0 > 3x0 = [[map(plus(s(0)), _x0)]] [[double(_x0)]] = 3 + 3x0 > 3x0 = [[map(times(s(s(0))), _x0)]] [[map(_F0, nil)]] = 6 + 2F0(2) > 2 = [[nil]] We can thus remove the following rules: inc(X) => map(plus(s(0)), X) double(X) => map(times(s(s(0))), X) map(F, nil) => nil We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): times(0, X) >? 0 We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 times = \y0y1.3 + y1 + 3y0 Using this interpretation, the requirements translate to: [[times(0, _x0)]] = 3 + x0 > 0 = [[0]] We can thus remove the following rules: times(0, X) => 0 All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.