We consider the system AotoYamada_05__Ex1SimplyTyped. Alphabet: 0 : [] --> a add : [a] --> a -> a cons : [b * c] --> c id : [] --> a -> a map : [b -> b * c] --> c nil : [] --> c s : [a] --> a Rules: id x => x add(0) => id add(s(x)) y => s(add(x) y) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, 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: id X => 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) QTRS Reverse [EQUIVALENT] || (2) QTRS || (3) RFCMatchBoundsTRSProof [EQUIVALENT] || (4) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || id(%X) -> %X || || Q is empty. || || ---------------------------------------- || || (1) QTRS Reverse (EQUIVALENT) || We applied the QTRS Reverse Processor [REVERSE]. || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || id(x) -> x || || Q is empty. || || ---------------------------------------- || || (3) RFCMatchBoundsTRSProof (EQUIVALENT) || Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R. || The following rules were used to construct the certificate: || || id(x) -> x || || The certificate found is represented by the following graph. || The certificate consists of the following enumerated nodes: || 3, 4 || || Node 3 is start node and node 4 is final node. || || Those nodes are connected through the following edges: || || * 3 to 4 labelled id_1(0), id_1(1)* 4 to 4 labelled #_1(0) || || || ---------------------------------------- || || (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] add(0) X =#> id X 1] add#(0) =#> id# 2] add(s(X)) Y =#> add(X) Y 3] add(s(X)) Y =#> add#(X) 4] map#(F, cons(X, Y)) =#> F(X) 5] map#(F, cons(X, Y)) =#> map#(F, Y) Rules R_0: id X => X add(0) => id add(s(X)) Y => s(add(X) Y) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(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 : * 1 : * 2 : 0, 2, 3 * 3 : 1 * 4 : 0, 1, 2, 3, 4, 5 * 5 : 4, 5 This graph has the following strongly connected components: P_1: add(s(X)) Y =#> add(X) Y P_2: map#(F, cons(X, Y)) =#> F(X) map#(F, cons(X, Y)) =#> map#(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) and (P_2, R_0, m, f). 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_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_2, R_0, minimal, formative) by (P_2, R_1, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (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 = \y0y1.1 + y0 + 2y1 map = \G0y1.y1 + 2y1G0(y1) map# = \G0y1.3 + 2y1 + G0(y1) Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 5 + 2x1 + 4x2 + F0(1 + x1 + 2x2) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 5 + 2x1 + 4x2 + F0(1 + x1 + 2x2) > 3 + 2x2 + F0(x2) = [[map#(_F0, _x2)]] [[map(_F0, cons(_x1, _x2))]] = 1 + x1 + 2x2 + 2x1F0(1 + x1 + 2x2) + 2F0(1 + x1 + 2x2) + 4x2F0(1 + x1 + 2x2) >= 1 + 2x2 + 4x2F0(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 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. 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(add) = 1 Thus, we can orient the dependency pairs as follows: nu(add(s(X)) Y) = s(X) |> X = nu(add(X) Y) By [FuhKop19, Thm. 61], 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 +++ [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.