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 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] 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)) =#> F(X) 19] map#(F, add(X, Y)) =#> map#(F, Y) 20] filter#(F, add(X, Y)) =#> filter2#(F X, F, X, Y) 21] filter#(F, add(X, Y)) =#> F(X) 22] filter2#(true, F, X, Y) =#> filter#(F, Y) 23] 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, 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 : 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 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23 * 19 : 18, 19 * 20 : 22, 23 * 21 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23 * 22 : 20, 21 * 23 : 20, 21 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)) =#> F(X) map#(F, add(X, Y)) =#> map#(F, Y) filter#(F, add(X, Y)) =#> filter2#(F X, F, X, Y) filter#(F, add(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). 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, add(X, Y)) >? F(X) map#(F, add(X, Y)) >? map#(F, Y) filter#(F, add(X, Y)) >? filter2#(F X, F, X, Y) filter#(F, add(X, Y)) >? F(X) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) 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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 add = \y0y1.1 + y1 + 2y0 app = \y0y1.y0 + y1 eq = \y0y1.2 false = 0 filter = \G0y1.y1 + G0(0) + G0(y1) + 2y1G0(y1) filter2 = \y0G1y2y3.1 + y3 + 2y2 + G1(0) + G1(y3) + 2y3G1(y3) filter2# = \y0G1y2y3.1 + y3G1(y3) filter# = \G0y1.1 + y1G0(y1) if!fac6220min = \y0y1.2y1 if!fac6220minsort = \y0y1y2.2y1 + 2y2 if!fac6220rm = \y0y1y2.y1 + y2 le = \y0y1.y1 map = \G0y1.3 + y1 + 2y1G0(y1) + 2G0(y1) map# = \G0y1.3 + G0(y1) min = \y0.2y0 minsort = \y0y1.2y0 + 2y1 nil = 0 rm = \y0y1.y0 + y1 s = \y0.3 + 2y0 true = 0 Using this interpretation, the requirements translate to: [[map#(_F0, add(_x1, _x2))]] = 3 + F0(1 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, add(_x1, _x2))]] = 3 + F0(1 + x2 + 2x1) >= 3 + F0(x2) = [[map#(_F0, _x2)]] [[filter#(_F0, add(_x1, _x2))]] = 1 + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) >= 1 + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, add(_x1, _x2))]] = 1 + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = 1 + x2F0(x2) >= 1 + x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 1 + x2F0(x2) >= 1 + x2F0(x2) = [[filter#(_F0, _x2)]] [[eq(0, 0)]] = 2 >= 0 = [[true]] [[eq(0, s(_x0))]] = 2 >= 0 = [[false]] [[eq(s(_x0), 0)]] = 2 >= 0 = [[false]] [[eq(s(_x0), s(_x1))]] = 2 >= 2 = [[eq(_x0, _x1)]] [[le(0, _x0)]] = x0 >= 0 = [[true]] [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] [[le(s(_x0), s(_x1))]] = 3 + 2x1 >= x1 = [[le(_x0, _x1)]] [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] [[app(add(_x0, _x1), _x2)]] = 1 + x1 + x2 + 2x0 >= 1 + x1 + x2 + 2x0 = [[add(_x0, app(_x1, _x2))]] [[min(add(_x0, nil))]] = 2 + 4x0 >= x0 = [[_x0]] [[min(add(_x0, add(_x1, _x2)))]] = 4 + 2x2 + 4x0 + 4x1 >= 4 + 2x2 + 4x0 + 4x1 = [[if!fac6220min(le(_x0, _x1), add(_x0, add(_x1, _x2)))]] [[if!fac6220min(true, add(_x0, add(_x1, _x2)))]] = 4 + 2x2 + 4x0 + 4x1 >= 2 + 2x2 + 4x0 = [[min(add(_x0, _x2))]] [[if!fac6220min(false, add(_x0, add(_x1, _x2)))]] = 4 + 2x2 + 4x0 + 4x1 >= 2 + 2x2 + 4x1 = [[min(add(_x1, _x2))]] [[rm(_x0, nil)]] = x0 >= 0 = [[nil]] [[rm(_x0, add(_x1, _x2))]] = 1 + x0 + x2 + 2x1 >= 1 + x0 + x2 + 2x1 = [[if!fac6220rm(eq(_x0, _x1), _x0, add(_x1, _x2))]] [[if!fac6220rm(true, _x0, add(_x1, _x2))]] = 1 + x0 + x2 + 2x1 >= x0 + x2 = [[rm(_x0, _x2)]] [[if!fac6220rm(false, _x0, add(_x1, _x2))]] = 1 + x0 + x2 + 2x1 >= 1 + x0 + x2 + 2x1 = [[add(_x1, rm(_x0, _x2))]] [[minsort(nil, nil)]] = 0 >= 0 = [[nil]] [[minsort(add(_x0, _x1), _x2)]] = 2 + 2x1 + 2x2 + 4x0 >= 2 + 2x1 + 2x2 + 4x0 = [[if!fac6220minsort(eq(_x0, min(add(_x0, _x1))), add(_x0, _x1), _x2)]] [[if!fac6220minsort(true, add(_x0, _x1), _x2)]] = 2 + 2x1 + 2x2 + 4x0 >= 1 + 2x1 + 2x2 + 4x0 = [[add(_x0, minsort(app(rm(_x0, _x1), _x2), nil))]] [[if!fac6220minsort(false, add(_x0, _x1), _x2)]] = 2 + 2x1 + 2x2 + 4x0 >= 2 + 2x1 + 2x2 + 4x0 = [[minsort(_x1, add(_x0, _x2))]] [[map(_F0, nil)]] = 3 + 2F0(0) >= 0 = [[nil]] [[map(_F0, add(_x1, _x2))]] = 4 + x2 + 2x1 + 2x2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) + 4F0(1 + x2 + 2x1) >= 4 + x2 + 2x2F0(x2) + 2F0(x2) + 2max(x1, F0(x1)) = [[add(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] [[filter(_F0, add(_x1, _x2))]] = 1 + x2 + 2x1 + F0(0) + 2x2F0(1 + x2 + 2x1) + 3F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 1 + x2 + 2x1 + F0(0) + F0(x2) + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 1 + x2 + 2x1 + F0(0) + F0(x2) + 2x2F0(x2) >= 1 + x2 + 2x1 + F0(0) + F0(x2) + 2x2F0(x2) = [[add(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 1 + x2 + 2x1 + F0(0) + F0(x2) + 2x2F0(x2) >= x2 + F0(0) + F0(x2) + 2x2F0(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_0, minimal, formative) by (P_8, R_0, minimal, formative), where P_8 consists of: map#(F, add(X, Y)) =#> map#(F, Y) 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) 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_0, minimal, formative) is finite. We consider the dependency pair problem (P_8, 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, 3 * 2 : 1 * 3 : 1 This graph has the following strongly connected components: P_9: map#(F, add(X, Y)) =#> map#(F, Y) P_10: 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_8, R_0, m, f) by (P_9, R_0, m, f) and (P_10, 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), (P_9, R_0, minimal, formative) and (P_10, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_10, R_0, minimal, 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 [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_10, R_0, minimal, f) by (P_11, R_0, minimal, f), where P_11 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, 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), (P_9, R_0, minimal, formative) and (P_11, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_11, 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 : * 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), (P_6, R_0, minimal, formative) and (P_9, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_9, R_0, minimal, 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 [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_9, 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), (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). 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, minimal, formative) by (P_12, R_0, minimal, formative), where P_12 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, 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_12, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_12, R_0, minimal, 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 [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_12, R_0, minimal, f) by (P_13, R_0, minimal, f), where P_13 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, 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_13, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_13, 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 : 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) and (P_5, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_0, minimal, 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 [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_0, minimal, f) by (P_14, R_0, minimal, f), where P_14 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, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_14, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_14, 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 : 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) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_0, minimal, 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, 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(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 [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). 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 [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, 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 (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(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 [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.