We consider the system Applicative_05__TreeHeight. Alphabet: 0 : [] --> d cons : [d * c] --> c false : [] --> a height : [] --> d -> d if : [a * d] --> d le : [d * d] --> a map : [d -> d * c] --> c maxlist : [d * c] --> d nil : [] --> c node : [b * c] --> d s : [d] --> d true : [] --> a Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) maxlist(x, cons(y, z)) => if(le(x, y), maxlist(y, z)) maxlist(x, nil) => x height node(x, y) => s(maxlist(0, map(height, y))) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We observe that the rules contain a first-order subset: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) maxlist(X, cons(Y, Z)) => if(le(X, Y), maxlist(Y, Z)) maxlist(X, nil) => X Moreover, the system is orthogonal. Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed terminating: || proof of resources/system.trs || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 || || || Termination w.r.t. Q of the given QTRS could be proven: || || (0) QTRS || (1) QTRSRRRProof [EQUIVALENT] || (2) QTRS || (3) QTRSRRRProof [EQUIVALENT] || (4) QTRS || (5) RisEmptyProof [EQUIVALENT] || (6) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || maxlist(%X, cons(%Y, %Z)) -> if(le(%X, %Y), maxlist(%Y, %Z)) || maxlist(%X, nil) -> %X || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(0) = 1 || POL(cons(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 || POL(false) = 1 || POL(if(x_1, x_2)) = 2 + 2*x_1 + x_2 || POL(le(x_1, x_2)) = x_1 + x_2 || POL(maxlist(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 || POL(nil) = 1 || POL(s(x_1)) = 1 + 2*x_1 || POL(true) = 0 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || maxlist(%X, nil) -> %X || || || || || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || maxlist(%X, cons(%Y, %Z)) -> if(le(%X, %Y), maxlist(%Y, %Z)) || || Q is empty. || || ---------------------------------------- || || (3) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(cons(x_1, x_2)) = 2 + 2*x_1 + x_2 || POL(if(x_1, x_2)) = 1 + x_1 + x_2 || POL(le(x_1, x_2)) = 2 + x_1 + 2*x_2 || POL(maxlist(x_1, x_2)) = 2*x_1 + 2*x_2 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || maxlist(%X, cons(%Y, %Z)) -> if(le(%X, %Y), maxlist(%Y, %Z)) || || || || || ---------------------------------------- || || (4) || Obligation: || Q restricted rewrite system: || R is empty. || Q is empty. || || ---------------------------------------- || || (5) RisEmptyProof (EQUIVALENT) || The TRS R is empty. Hence, termination is trivially proven. || ---------------------------------------- || || (6) || YES || 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] map#(F, cons(X, Y)) =#> F(X) 1] map#(F, cons(X, Y)) =#> map#(F, Y) 2] height node(X, Y) =#> maxlist#(0, map(height, Y)) 3] height node(X, Y) =#> map#(height, Y) 4] height node(X, Y) =#> height# Rules R_0: map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) maxlist(X, cons(Y, Z)) => if(le(X, Y), maxlist(Y, Z)) maxlist(X, nil) => X height node(X, Y) => s(maxlist(0, map(height, 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, 3, 4 * 1 : 0, 1 * 2 : * 3 : 0, 1 * 4 : This graph has the following strongly connected components: P_1: map#(F, cons(X, Y)) =#> F(X) map#(F, cons(X, Y)) =#> map#(F, Y) height node(X, Y) =#> map#(height, 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). 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 R_1 ::= map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) maxlist(X, nil) => X 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 [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) height node(X, Y) >? map#(height, Y) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) maxlist(X, nil) >= X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.1 + y0 + y1 height = \y0.0 map = \G0y1.2 + y1 + G0(0) + 3y1G0(y1) map# = \G0y1.1 + 2G0(y1) maxlist = \y0y1.y0 nil = 0 node = \y0y1.3 + 2y1 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 1 + 2F0(1 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 1 + 2F0(1 + x1 + x2) >= 1 + 2F0(x2) = [[map#(_F0, _x2)]] [[height node(_x0, _x1)]] = 3 + 2x1 > 1 = [[map#(height, _x1)]] [[map(_F0, nil)]] = 2 + F0(0) >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + F0(0) + 3x1F0(1 + x1 + x2) + 3x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) >= 3 + x2 + F0(0) + 3x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[maxlist(_x0, nil)]] = x0 >= x0 = [[_x0]] 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_2, R_1, minimal, formative), where P_2 consists of: map#(F, cons(X, Y)) =#> map#(F, Y) Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, 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 [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_1, minimal, f) by ({}, R_1, 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 +++ [FuhKop19] C. Fuhs, and C. Kop. A static higher-order dependency pair framework. In Proceedings of ESOP 2019, 2019. [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.