We consider the system Applicative_first_order_05__#3.57. Alphabet: 0 : [] --> b app : [c * c] --> c cons : [b * c] --> c false : [] --> a filter : [b -> a * c] --> c filter2 : [a * b -> a * b * c] --> c map : [b -> b * c] --> c minus : [b * b] --> b nil : [] --> c plus : [b * b] --> b quot : [b * b] --> b s : [b] --> b sum : [c] --> c true : [] --> a Rules: minus(x, 0) => x minus(s(x), s(y)) => minus(x, y) minus(minus(x, y), z) => minus(x, plus(y, z)) quot(0, s(x)) => 0 quot(s(x), s(y)) => s(quot(minus(x, y), s(y))) plus(0, x) => x plus(s(x), y) => s(plus(x, y)) app(nil, x) => x app(x, nil) => x app(cons(x, y), z) => cons(x, app(y, z)) sum(cons(x, nil)) => cons(x, nil) sum(cons(x, cons(y, z))) => sum(cons(plus(x, y), z)) sum(app(x, cons(y, cons(z, u)))) => sum(app(x, sum(cons(y, cons(z, u))))) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) filter(f, nil) => nil filter(f, cons(x, y)) => filter2(f x, f, x, y) filter2(true, f, x, y) => cons(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 dynamic dependency pairs. After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] minus#(s(X), s(Y)) =#> minus#(X, Y) 1] minus#(minus(X, Y), Z) =#> minus#(X, plus(Y, Z)) 2] minus#(minus(X, Y), Z) =#> plus#(Y, Z) 3] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) 4] quot#(s(X), s(Y)) =#> minus#(X, Y) 5] plus#(s(X), Y) =#> plus#(X, Y) 6] app#(cons(X, Y), Z) =#> app#(Y, Z) 7] sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z)) 8] sum#(cons(X, cons(Y, Z))) =#> plus#(X, Y) 9] sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U))))) 10] sum#(app(X, cons(Y, cons(Z, U)))) =#> app#(X, sum(cons(Y, cons(Z, U)))) 11] sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(cons(Y, cons(Z, U))) 12] map#(F, cons(X, Y)) =#> F(X) 13] map#(F, cons(X, Y)) =#> map#(F, Y) 14] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 15] filter#(F, cons(X, Y)) =#> F(X) 16] filter2#(true, F, X, Y) =#> filter#(F, Y) 17] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, 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, 2 * 1 : 0, 1, 2 * 2 : 5 * 3 : 3, 4 * 4 : 0, 1, 2 * 5 : 5 * 6 : 6 * 7 : 7, 8 * 8 : 5 * 9 : 7, 8, 9, 10, 11 * 10 : 6 * 11 : 7, 8 * 12 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 13 : 12, 13 * 14 : 16, 17 * 15 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 16 : 14, 15 * 17 : 14, 15 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) minus#(minus(X, Y), Z) =#> minus#(X, plus(Y, Z)) P_2: quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) P_3: plus#(s(X), Y) =#> plus#(X, Y) P_4: app#(cons(X, Y), Z) =#> app#(Y, Z) P_5: sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z)) P_6: sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U))))) P_7: map#(F, cons(X, Y)) =#> F(X) map#(F, cons(X, Y)) =#> map#(F, Y) filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) =#> F(X) 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) and (P_7, 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), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative) and (P_7, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_7, R_0, minimal, formative). The formative rules of (P_7, R_0) are R_1 ::= app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_7, R_0, minimal, formative) by (P_7, 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), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative) and (P_7, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_7, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70]. Thus, we must orient: map#(F, cons(X, Y)) >? F(X) map#(F, cons(X, Y)) >? map#(F, Y) filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) >? F(X) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) app(nil, X) >= X app(X, nil) >= X app(cons(X, Y), Z) >= cons(X, app(Y, Z)) sum(cons(X, nil)) >= cons(X, nil) sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(Z, U))))) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: app = \y0y1.y1 + 3y0 cons = \y0y1.2 + y0 + y1 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.2 + y2 + y3 filter2# = \y0G1y2y3.2y3 + G1(y3) filter# = \G0y1.2y1 + G0(y1) map = \G0y1.y1 + y1G0(y1) map# = \G0y1.3 + y1 + G0(y1) nil = 0 plus = \y0y1.0 sum = \y0.y0 true = 3 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 5 + x1 + x2 + F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 5 + x1 + x2 + F0(2 + x1 + x2) > 3 + x2 + F0(x2) = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(2 + x1 + x2) > 2x2 + F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = 2x2 + F0(x2) >= 2x2 + F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 2x2 + F0(x2) >= 2x2 + F0(x2) = [[filter#(_F0, _x2)]] [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] [[app(cons(_x0, _x1), _x2)]] = 6 + x2 + 3x0 + 3x1 >= 2 + x0 + x2 + 3x1 = [[cons(_x0, app(_x1, _x2))]] [[sum(cons(_x0, nil))]] = 2 + x0 >= 2 + x0 = [[cons(_x0, nil)]] [[sum(cons(_x0, cons(_x1, _x2)))]] = 4 + x0 + x1 + x2 >= 2 + x2 = [[sum(cons(plus(_x0, _x1), _x2))]] [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 4 + x1 + x2 + x3 + 3x0 >= 4 + x1 + x2 + x3 + 3x0 = [[sum(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 2 + x2 + x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + x1 + x2 >= x2 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_7, R_1, minimal, formative) by (P_8, R_1, minimal, formative), where P_8 consists of: 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, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative) and (P_8, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_8, R_1, 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 : * 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, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative) and (P_6, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_6, R_0, minimal, formative). The formative rules of (P_6, R_0) are exactly R_1: app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_6, R_0, minimal, formative) by (P_6, 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), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative) and (P_6, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_6, R_1, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_6, R_1) are: app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) app(nil, X) >= X app(X, nil) >= X app(cons(X, Y), Z) >= cons(X, app(Y, Z)) sum(cons(X, nil)) >= cons(X, nil) sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(Z, U))))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: app = \y0y1.y1 + 3y0 cons = \y0y1.2 + y1 nil = 0 plus = \y0y1.0 sum = \y0.2 sum# = \y0.2y0 Using this interpretation, the requirements translate to: [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 8 + 2x3 + 6x0 > 4 + 6x0 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] [[app(_x0, nil)]] = 3x0 >= x0 = [[_x0]] [[app(cons(_x0, _x1), _x2)]] = 6 + x2 + 3x1 >= 2 + x2 + 3x1 = [[cons(_x0, app(_x1, _x2))]] [[sum(cons(_x0, nil))]] = 2 >= 2 = [[cons(_x0, nil)]] [[sum(cons(_x0, cons(_x1, _x2)))]] = 2 >= 2 = [[sum(cons(plus(_x0, _x1), _x2))]] [[sum(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 2 >= 2 = [[sum(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_6, 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), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_0, minimal, formative). The formative rules of (P_5, R_0) are exactly R_1: app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_5, R_0, minimal, formative) by (P_5, 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), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_5, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_1, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. (P_5, R_1) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: sum#(cons(X, cons(Y, Z))) >? sum#(cons(plus(X, Y), Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.3 + 3y1 plus = \y0y1.0 sum# = \y0.y0 Using this interpretation, the requirements translate to: [[sum#(cons(_x0, cons(_x1, _x2)))]] = 12 + 9x2 > 3 + 3x2 = [[sum#(cons(plus(_x0, _x1), _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_5, 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), (P_2, R_0, minimal, formative), (P_3, 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 apply the subterm criterion with the following projection function: nu(app#) = 1 Thus, we can orient the dependency pairs as follows: nu(app#(cons(X, Y), Z)) = cons(X, Y) |> Y = nu(app#(Y, Z)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_0, minimal, f) by ({}, R_0, minimal, 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, 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). We apply the subterm criterion with the following projection function: nu(plus#) = 1 Thus, we can orient the dependency pairs as follows: nu(plus#(s(X), Y)) = s(X) |> X = nu(plus#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by ({}, R_0, minimal, 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, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). The formative rules of (P_2, R_0) are R_2 ::= minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_2, R_2, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_2, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_2, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_2, R_2) are: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: quot#(s(X), s(Y)) >? quot#(minus(X, Y), s(Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 minus = \y0y1.y0 plus = \y0y1.y1 + 3y0 quot# = \y0y1.2y0 s = \y0.3 + y0 Using this interpretation, the requirements translate to: [[quot#(s(_x0), s(_x1))]] = 6 + 2x0 > 2x0 = [[quot#(minus(_x0, _x1), s(_x1))]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = 3 + x0 >= x0 = [[minus(_x0, _x1)]] [[minus(minus(_x0, _x1), _x2)]] = x0 >= x0 = [[minus(_x0, plus(_x1, _x2))]] [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = 9 + x1 + 3x0 >= 3 + x1 + 3x0 = [[s(plus(_x0, _x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_2) by ({}, R_2). 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, 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(minus#) = 1 Thus, we can orient the dependency pairs as follows: nu(minus#(s(X), s(Y))) = s(X) |> X = nu(minus#(X, Y)) nu(minus#(minus(X, Y), Z)) = minus(X, Y) |> X = nu(minus#(X, plus(Y, Z))) 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 +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.