We consider the system Applicative_05__TreeSize. Alphabet: 0 : [] --> c cons : [c * b] --> b map : [c -> c * b] --> b nil : [] --> b node : [a * b] --> c plus : [c * c] --> c s : [c] --> c size : [] --> c -> c sum : [b] --> c Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) sum(cons(x, y)) => plus(x, sum(y)) size node(x, y) => s(sum(map(size, y))) plus(0, x) => 0 plus(s(x), y) => s(plus(x, 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: sum(cons(X, Y)) => plus(X, sum(Y)) plus(0, X) => 0 plus(s(X), Y) => s(plus(X, Y)) 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) RisEmptyProof [EQUIVALENT] || (4) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || sum(cons(%X, %Y)) -> plus(%X, sum(%Y)) || plus(0, %X) -> 0 || plus(s(%X), %Y) -> s(plus(%X, %Y)) || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(0) = 1 || POL(cons(x_1, x_2)) = 2 + x_1 + 2*x_2 || POL(plus(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 || POL(s(x_1)) = 1 + x_1 || POL(sum(x_1)) = 1 + 2*x_1 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || sum(cons(%X, %Y)) -> plus(%X, sum(%Y)) || plus(0, %X) -> 0 || plus(s(%X), %Y) -> s(plus(%X, %Y)) || || || || || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || R is empty. || Q is empty. || || ---------------------------------------- || || (3) RisEmptyProof (EQUIVALENT) || The TRS R is empty. Hence, termination is trivially proven. || ---------------------------------------- || || (4) || 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] size node(X, Y) =#> sum#(map(size, Y)) 3] size node(X, Y) =#> map#(size, Y) 4] size node(X, Y) =#> size# Rules R_0: map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) sum(cons(X, Y)) => plus(X, sum(Y)) size node(X, Y) => s(sum(map(size, Y))) plus(0, X) => 0 plus(s(X), Y) => s(plus(X, 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) size node(X, Y) =#> map#(size, 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, cons(X, Y)) => cons(F X, map(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 [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) size node(X, Y) >? map#(size, Y) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.2 + y0 + 2y1 map = \G0y1.y1 + y1G0(y1) map# = \G0y1.2G0(y1) node = \y0y1.3 size = \y0.0 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 2F0(2 + x1 + 2x2) >= F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 2F0(2 + x1 + 2x2) >= 2F0(x2) = [[map#(_F0, _x2)]] [[size node(_x0, _x1)]] = 3 > 0 = [[map#(size, _x1)]] [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + 2x2 + 2x2F0(2 + x1 + 2x2) + 2F0(2 + x1 + 2x2) + x1F0(2 + x1 + 2x2) >= 2 + 2x2 + 2x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _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_2, R_1, minimal, formative), where P_2 consists of: map#(F, cons(X, Y)) =#> F(X) 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 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) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.2 + y1 + 2y0 map = \G0y1.y1 + 2y1G0(y1) map# = \G0y1.3 + y1 + y1G0(y1) Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > 3 + x2 + x2F0(x2) = [[map#(_F0, _x2)]] [[map(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) >= 2 + x2 + 2x2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_1) by ({}, R_1). 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.