We consider the system Applicative_AG01_innermost__#4.28. Alphabet: 0 : [] --> a bits : [a] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d half : [a] --> a map : [c -> c * d] --> d nil : [] --> d s : [a] --> a true : [] --> b Rules: half(0) => 0 half(s(0)) => 0 half(s(s(x))) => s(half(x)) bits(0) => 0 bits(s(x)) => s(bits(half(s(x)))) 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 observe that the rules contain a first-order subset: half(0) => 0 half(s(0)) => 0 half(s(s(X))) => s(half(X)) bits(0) => 0 bits(s(X)) => s(bits(half(s(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: || || half(0) -> 0 || half(s(0)) -> 0 || half(s(s(%X))) -> s(half(%X)) || bits(0) -> 0 || bits(s(%X)) -> s(bits(half(s(%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: || || 0'(half(x)) -> 0'(x) || 0'(s(half(x))) -> 0'(x) || s(s(half(x))) -> half(s(x)) || 0'(bits(x)) -> 0'(x) || s(bits(x)) -> s(half(bits(s(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: || || 0'(half(x)) -> 0'(x) || 0'(s(half(x))) -> 0'(x) || s(s(half(x))) -> half(s(x)) || 0'(bits(x)) -> 0'(x) || s(bits(x)) -> s(half(bits(s(x)))) || || The certificate found is represented by the following graph. || The certificate consists of the following enumerated nodes: || 1, 2, 5, 6, 7, 8, 9, 10, 11, 12 || || Node 1 is start node and node 2 is final node. || || Those nodes are connected through the following edges: || || * 1 to 2 labelled 0'_1(0), 0'_1(1)* 1 to 5 labelled half_1(0)* 1 to 6 labelled s_1(0)* 2 to 2 labelled #_1(0)* 5 to 2 labelled s_1(0)* 5 to 9 labelled half_1(1)* 5 to 10 labelled s_1(1)* 6 to 7 labelled half_1(0)* 7 to 8 labelled bits_1(0)* 8 to 2 labelled s_1(0)* 8 to 9 labelled half_1(1)* 8 to 10 labelled s_1(1)* 9 to 2 labelled s_1(1)* 9 to 9 labelled half_1(1)* 9 to 10 labelled s_1(1)* 10 to 11 labelled half_1(1)* 11 to 12 labelled bits_1(1)* 12 to 2 labelled s_1(1)* 12 to 9 labelled half_1(1)* 12 to 10 labelled s_1(1) || || || ---------------------------------------- || || (4) || YES || We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [FuhKop19]). We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] map#(F, cons(X, Y)) =#> map#(F, Y) 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 3] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: half(0) => 0 half(s(0)) => 0 half(s(s(X))) => s(half(X)) bits(0) => 0 bits(s(X)) => s(bits(half(s(X)))) 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, computable, formative) is finite. We consider the dependency pair problem (P_0, R_0, computable, 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_1: map#(F, cons(X, Y)) =#> map#(F, Y) P_2: filter#(F, cons(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_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, computable, formative) and (P_2, R_0, computable, formative) is finite. We consider the dependency pair problem (P_2, R_0, computable, 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, cons(X, Y))) = cons(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 [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, computable, f) by (P_3, R_0, computable, f), where P_3 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, computable, formative) and (P_3, R_0, computable, formative) is finite. We consider the dependency pair problem (P_3, R_0, computable, 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 (P_1, R_0, computable, formative) is finite. We consider the dependency pair problem (P_1, R_0, computable, 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_1, R_0, computable, f) by ({}, R_0, computable, 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. [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009.