We consider the system Applicative_first_order_05__#3.38. Alphabet: 0 : [] --> c cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d nil : [] --> d rev : [d] --> d rev1 : [c * d] --> c rev2 : [c * d] --> d s : [a] --> c true : [] --> b Rules: rev(nil) => nil rev(cons(x, y)) => cons(rev1(x, y), rev2(x, y)) rev1(0, nil) => 0 rev1(s(x), nil) => s(x) rev1(x, cons(y, z)) => rev1(y, z) rev2(x, nil) => nil rev2(x, cons(y, z)) => rev(cons(x, rev2(y, z))) 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] rev#(cons(X, Y)) =#> rev1#(X, Y) 1] rev#(cons(X, Y)) =#> rev2#(X, Y) 2] rev1#(X, cons(Y, Z)) =#> rev1#(Y, Z) 3] rev2#(X, cons(Y, Z)) =#> rev#(cons(X, rev2(Y, Z))) 4] rev2#(X, cons(Y, Z)) =#> rev2#(Y, Z) 5] map#(F, cons(X, Y)) =#> F(X) 6] map#(F, cons(X, Y)) =#> map#(F, Y) 7] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 8] filter#(F, cons(X, Y)) =#> F(X) 9] filter2#(true, F, X, Y) =#> filter#(F, Y) 10] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: rev(nil) => nil rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) rev1(0, nil) => 0 rev1(s(X), nil) => s(X) rev1(X, cons(Y, Z)) => rev1(Y, Z) rev2(X, nil) => nil rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) 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 : 2 * 1 : 3, 4 * 2 : 2 * 3 : 0, 1 * 4 : 3, 4 * 5 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 * 6 : 5, 6 * 7 : 9, 10 * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 * 9 : 7, 8 * 10 : 7, 8 This graph has the following strongly connected components: P_1: rev#(cons(X, Y)) =#> rev2#(X, Y) rev2#(X, cons(Y, Z)) =#> rev#(cons(X, rev2(Y, Z))) rev2#(X, cons(Y, Z)) =#> rev2#(Y, Z) P_2: rev1#(X, cons(Y, Z)) =#> rev1#(Y, Z) P_3: 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) and (P_3, 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) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative). The formative rules of (P_3, R_0) are R_1 ::= rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) map(F, cons(X, Y)) => cons(F X, map(F, Y)) 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_3, R_0, minimal, formative) by (P_3, 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) and (P_3, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_3, 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) rev(cons(X, Y)) >= cons(rev1(X, Y), rev2(X, Y)) rev2(X, cons(Y, Z)) >= rev(cons(X, rev2(Y, Z))) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 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: cons = \y0y1.1 + y1 + 2y0 false = 3 filter = \G0y1.y1 + G0(y1) + y1G0(y1) filter2 = \y0G1y2y3.1 + y3 + 2y2 + G1(y3) + y3G1(y3) filter2# = \y0G1y2y3.2y3G1(y3) filter# = \G0y1.2y1G0(y1) map = \G0y1.2y1 + 2y1G0(y1) map# = \G0y1.3 + y1G0(y1) rev = \y0.y0 rev1 = \y0y1.0 rev2 = \y0y1.y1 + 2y0 true = 3 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 3 + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 3 + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) >= 3 + x2F0(x2) = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2x2F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 2x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2x2F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = 2x2F0(x2) >= 2x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 2x2F0(x2) >= 2x2F0(x2) = [[filter#(_F0, _x2)]] [[rev(cons(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[cons(rev1(_x0, _x1), rev2(_x0, _x1))]] [[rev2(_x0, cons(_x1, _x2))]] = 1 + x2 + 2x0 + 2x1 >= 1 + x2 + 2x0 + 2x1 = [[rev(cons(_x0, rev2(_x1, _x2)))]] [[map(_F0, cons(_x1, _x2))]] = 2 + 2x2 + 4x1 + 2x2F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 1 + 2x2 + 2x2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 1 + x2 + 2x1 + 2x1F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) >= 1 + x2 + 2x1 + F0(x2) + x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 1 + x2 + 2x1 + F0(x2) + x2F0(x2) >= 1 + x2 + 2x1 + F0(x2) + x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 1 + x2 + 2x1 + F0(x2) + x2F0(x2) >= x2 + F0(x2) + x2F0(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_3, R_1, minimal, formative) by (P_4, R_1, minimal, formative), where P_4 consists of: 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) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_4, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_4, 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 : 0 * 1 : 3, 4 * 2 : 0, 1, 2, 3, 4 * 3 : 1, 2 * 4 : 1, 2 This graph has the following strongly connected components: P_5: map#(F, cons(X, Y)) =#> map#(F, Y) P_6: 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_4, R_1, m, f) by (P_5, R_1, m, f) and (P_6, R_1, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_5, R_1, 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 [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: 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) rev(cons(X, Y)) >= cons(rev1(X, Y), rev2(X, Y)) rev2(X, cons(Y, Z)) >= rev(cons(X, rev2(Y, Z))) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 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: cons = \y0y1.1 + y0 + y1 false = 3 filter = \G0y1.2y1 filter2 = \y0G1y2y3.2 + 2y2 + 2y3 filter2# = \y0G1y2y3.1 + y3 + y3G1(y3) filter# = \G0y1.y1 + y1G0(y1) map = \G0y1.2y1 + y1G0(y1) rev = \y0.y0 rev1 = \y0y1.0 rev2 = \y0y1.y0 + y1 true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) >= 1 + x2 + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = 1 + x2 + x2F0(x2) > x2 + x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 1 + x2 + x2F0(x2) > x2 + x2F0(x2) = [[filter#(_F0, _x2)]] [[rev(cons(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[cons(rev1(_x0, _x1), rev2(_x0, _x1))]] [[rev2(_x0, cons(_x1, _x2))]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[rev(cons(_x0, rev2(_x1, _x2)))]] [[map(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) >= 1 + 2x2 + x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 >= 2 + 2x1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + 2x1 + 2x2 >= 1 + x1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + 2x1 + 2x2 >= 2x2 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_6, R_1, minimal, formative) by (P_7, R_1, minimal, formative), where P_7 consists of: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_5, R_1, minimal, formative) and (P_7, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_7, 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 : 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) and (P_5, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_1, 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, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_1, minimal, f) by ({}, R_1, 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(rev1#) = 2 Thus, we can orient the dependency pairs as follows: nu(rev1#(X, cons(Y, Z))) = cons(Y, Z) |> Z = nu(rev1#(Y, Z)) 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). The formative rules of (P_1, R_0) are exactly R_1: rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) map(F, cons(X, Y)) => cons(F X, map(F, Y)) 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_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative). Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_1, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_1, R_1) are: rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: rev#(cons(X, Y)) >? rev2#(X, Y) rev2#(X, cons(Y, Z)) >? rev#(cons(X, rev2(Y, Z))) rev2#(X, cons(Y, Z)) >? rev2#(Y, Z) rev(cons(X, Y)) >= cons(rev1(X, Y), rev2(X, Y)) rev2(X, cons(Y, Z)) >= rev(cons(X, rev2(Y, Z))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.2 + y1 rev = \y0.y0 rev1 = \y0y1.0 rev2 = \y0y1.y1 rev2# = \y0y1.2y1 rev# = \y0.2y0 Using this interpretation, the requirements translate to: [[rev#(cons(_x0, _x1))]] = 4 + 2x1 > 2x1 = [[rev2#(_x0, _x1)]] [[rev2#(_x0, cons(_x1, _x2))]] = 4 + 2x2 >= 4 + 2x2 = [[rev#(cons(_x0, rev2(_x1, _x2)))]] [[rev2#(_x0, cons(_x1, _x2))]] = 4 + 2x2 > 2x2 = [[rev2#(_x1, _x2)]] [[rev(cons(_x0, _x1))]] = 2 + x1 >= 2 + x1 = [[cons(rev1(_x0, _x1), rev2(_x0, _x1))]] [[rev2(_x0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[rev(cons(_x0, rev2(_x1, _x2)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_8, R_1, minimal, formative), where P_8 consists of: rev2#(X, cons(Y, Z)) =#> rev#(cons(X, rev2(Y, Z))) Thus, the original system is terminating if (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 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 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.