We consider the system onearg. Alphabet: 0 : [] --> nat add : [nat] --> nat -> nat eq : [nat] --> nat -> bool err : [] --> nat false : [] --> bool id : [] --> nat -> nat nul : [] --> nat -> bool pred : [nat] --> nat s : [nat] --> nat true : [] --> bool Rules: nul 0 => true nul s(x) => false nul err => false pred(0) => err pred(s(x)) => x id x => x eq(0) => nul eq(s(x)) => /\y.eq(x) pred(y) add(0) => id add(s(x)) => /\y.add(x) s(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: nul 0 => true nul s(X) => false nul err => false pred(0) => err pred(s(X)) => X 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) RFCMatchBoundsTRSProof [EQUIVALENT] || (2) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || nul(0) -> true || nul(s(%X)) -> false || nul(err) -> false || pred(0) -> err || pred(s(%X)) -> %X || id(%X) -> %X || || Q is empty. || || ---------------------------------------- || || (1) 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: || || nul(0) -> true || nul(s(%X)) -> false || nul(err) -> false || pred(0) -> err || pred(s(%X)) -> %X || id(%X) -> %X || || The certificate found is represented by the following graph. || The certificate consists of the following enumerated nodes: || 2, 4 || || Node 2 is start node and node 4 is final node. || || Those nodes are connected through the following edges: || || * 2 to 4 labelled true(0), false(0), err(0), nul_1(0), 0(0), s_1(0), pred_1(0), id_1(0), true(1), false(1), err(1), nul_1(1), 0(1), s_1(1), pred_1(1), id_1(1)* 4 to 4 labelled #_1(0) || || || ---------------------------------------- || || (2) || YES || We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs. To start, the system is beta-saturated by adding the following rules: eq(s(X)) Y => eq(X) pred(Y) add(s(X)) Y => add(X) s(Y) We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] eq(0) X =#> nul X 1] eq#(0) =#> nul# 2] eq#(s(X)) =#> eq(X) pred(x) 3] eq#(s(X)) =#> pred#(x) 4] eq#(s(X)) =#> eq#(X) 5] add(0) X =#> id X 6] add#(0) =#> id# 7] add#(s(X)) =#> add(X) s(x) 8] add#(s(X)) =#> add#(X) 9] eq(s(X)) Y =#> eq(X) pred(Y) 10] eq(s(X)) Y =#> pred#(Y) 11] eq(s(X)) Y =#> eq#(X) 12] add(s(X)) Y =#> add(X) s(Y) 13] add(s(X)) Y =#> add#(X) Rules R_0: nul 0 => true nul s(X) => false nul err => false pred(0) => err pred(s(X)) => X id X => X eq(0) => nul eq(s(X)) => /\x.eq(X) pred(x) add(0) => id add(s(X)) => /\x.add(X) s(x) eq(s(X)) Y => eq(X) pred(Y) add(s(X)) Y => add(X) s(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 : * 3 : * 4 : 1, 2, 3, 4 * 5 : * 6 : * 7 : * 8 : 6, 7, 8 * 9 : 0, 9, 10, 11 * 10 : * 11 : 1, 2, 3, 4 * 12 : 5, 12, 13 * 13 : 6, 7, 8 This graph has the following strongly connected components: P_1: eq#(s(X)) =#> eq#(X) P_2: add#(s(X)) =#> add#(X) P_3: eq(s(X)) Y =#> eq(X) pred(Y) P_4: add(s(X)) Y =#> add(X) s(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), (P_3, R_0, m, f) and (P_4, 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), (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, 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) s(Y)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_4, 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 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). We apply the subterm criterion with the following projection function: nu(eq) = 1 Thus, we can orient the dependency pairs as follows: nu(eq(s(X)) Y) = s(X) |> X = nu(eq(X) pred(Y)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_3, 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 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(add#) = 1 Thus, we can orient the dependency pairs as follows: nu(add#(s(X))) = s(X) |> X = nu(add#(X)) By [FuhKop19, Thm. 61], 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). We apply the subterm criterion with the following projection function: nu(eq#) = 1 Thus, we can orient the dependency pairs as follows: nu(eq#(s(X))) = s(X) |> X = nu(eq#(X)) 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.