1.93/1.04 YES 1.93/1.05 We consider the system theBenchmark. 1.93/1.05 1.93/1.05 Alphabet: 1.93/1.05 1.93/1.05 0 : [] --> a 1.93/1.05 cons : [c * d] --> d 1.93/1.05 false : [] --> b 1.93/1.05 filter : [c -> b * d] --> d 1.93/1.05 filter2 : [b * c -> b * c * d] --> d 1.93/1.05 map : [c -> c * d] --> d 1.93/1.05 nil : [] --> d 1.93/1.05 quot : [a * a * a] --> a 1.93/1.05 s : [a] --> a 1.93/1.05 true : [] --> b 1.93/1.05 1.93/1.05 Rules: 1.93/1.05 1.93/1.05 quot(0, s(x), s(y)) => 0 1.93/1.05 quot(s(x), s(y), z) => quot(x, y, z) 1.93/1.05 quot(x, 0, s(y)) => s(quot(x, s(y), s(y))) 1.93/1.05 map(f, nil) => nil 1.93/1.05 map(f, cons(x, y)) => cons(f x, map(f, y)) 1.93/1.05 filter(f, nil) => nil 1.93/1.05 filter(f, cons(x, y)) => filter2(f x, f, x, y) 1.93/1.05 filter2(true, f, x, y) => cons(x, filter(f, y)) 1.93/1.05 filter2(false, f, x, y) => filter(f, y) 1.93/1.05 1.93/1.05 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 1.93/1.05 1.93/1.05 We observe that the rules contain a first-order subset: 1.93/1.05 1.93/1.05 quot(0, s(X), s(Y)) => 0 1.93/1.05 quot(s(X), s(Y), Z) => quot(X, Y, Z) 1.93/1.05 quot(X, 0, s(Y)) => s(quot(X, s(Y), s(Y))) 1.93/1.05 1.93/1.05 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. 1.93/1.05 1.93/1.05 According to the external first-order termination prover, this system is indeed terminating: 1.93/1.05 1.93/1.05 || proof of resources/system.trs 1.93/1.05 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 1.93/1.05 || 1.93/1.05 || 1.93/1.05 || Termination w.r.t. Q of the given QTRS could be proven: 1.93/1.05 || 1.93/1.05 || (0) QTRS 1.93/1.05 || (1) Overlay + Local Confluence [EQUIVALENT] 1.93/1.05 || (2) QTRS 1.93/1.05 || (3) DependencyPairsProof [EQUIVALENT] 1.93/1.05 || (4) QDP 1.93/1.05 || (5) UsableRulesProof [EQUIVALENT] 1.93/1.05 || (6) QDP 1.93/1.05 || (7) QReductionProof [EQUIVALENT] 1.93/1.05 || (8) QDP 1.93/1.05 || (9) QDPSizeChangeProof [EQUIVALENT] 1.93/1.05 || (10) YES 1.93/1.05 || 1.93/1.05 || 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (0) 1.93/1.05 || Obligation: 1.93/1.05 || Q restricted rewrite system: 1.93/1.05 || The TRS R consists of the following rules: 1.93/1.05 || 1.93/1.05 || quot(0, s(%X), s(%Y)) -> 0 1.93/1.05 || quot(s(%X), s(%Y), %Z) -> quot(%X, %Y, %Z) 1.93/1.05 || quot(%X, 0, s(%Y)) -> s(quot(%X, s(%Y), s(%Y))) 1.93/1.05 || 1.93/1.05 || Q is empty. 1.93/1.05 || 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (1) Overlay + Local Confluence (EQUIVALENT) 1.93/1.05 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (2) 1.93/1.05 || Obligation: 1.93/1.05 || Q restricted rewrite system: 1.93/1.05 || The TRS R consists of the following rules: 1.93/1.05 || 1.93/1.05 || quot(0, s(%X), s(%Y)) -> 0 1.93/1.05 || quot(s(%X), s(%Y), %Z) -> quot(%X, %Y, %Z) 1.93/1.05 || quot(%X, 0, s(%Y)) -> s(quot(%X, s(%Y), s(%Y))) 1.93/1.05 || 1.93/1.05 || The set Q consists of the following terms: 1.93/1.05 || 1.93/1.05 || quot(0, s(x0), s(x1)) 1.93/1.05 || quot(s(x0), s(x1), x2) 1.93/1.05 || quot(x0, 0, s(x1)) 1.93/1.05 || 1.93/1.05 || 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (3) DependencyPairsProof (EQUIVALENT) 1.93/1.05 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (4) 1.93/1.05 || Obligation: 1.93/1.05 || Q DP problem: 1.93/1.05 || The TRS P consists of the following rules: 1.93/1.05 || 1.93/1.05 || QUOT(s(%X), s(%Y), %Z) -> QUOT(%X, %Y, %Z) 1.93/1.05 || QUOT(%X, 0, s(%Y)) -> QUOT(%X, s(%Y), s(%Y)) 1.93/1.05 || 1.93/1.05 || The TRS R consists of the following rules: 1.93/1.05 || 1.93/1.05 || quot(0, s(%X), s(%Y)) -> 0 1.93/1.05 || quot(s(%X), s(%Y), %Z) -> quot(%X, %Y, %Z) 1.93/1.05 || quot(%X, 0, s(%Y)) -> s(quot(%X, s(%Y), s(%Y))) 1.93/1.05 || 1.93/1.05 || The set Q consists of the following terms: 1.93/1.05 || 1.93/1.05 || quot(0, s(x0), s(x1)) 1.93/1.05 || quot(s(x0), s(x1), x2) 1.93/1.05 || quot(x0, 0, s(x1)) 1.93/1.05 || 1.93/1.05 || We have to consider all minimal (P,Q,R)-chains. 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (5) UsableRulesProof (EQUIVALENT) 1.93/1.05 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (6) 1.93/1.05 || Obligation: 1.93/1.05 || Q DP problem: 1.93/1.05 || The TRS P consists of the following rules: 1.93/1.05 || 1.93/1.05 || QUOT(s(%X), s(%Y), %Z) -> QUOT(%X, %Y, %Z) 1.93/1.05 || QUOT(%X, 0, s(%Y)) -> QUOT(%X, s(%Y), s(%Y)) 1.93/1.05 || 1.93/1.05 || R is empty. 1.93/1.05 || The set Q consists of the following terms: 1.93/1.05 || 1.93/1.05 || quot(0, s(x0), s(x1)) 1.93/1.05 || quot(s(x0), s(x1), x2) 1.93/1.05 || quot(x0, 0, s(x1)) 1.93/1.05 || 1.93/1.05 || We have to consider all minimal (P,Q,R)-chains. 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (7) QReductionProof (EQUIVALENT) 1.93/1.05 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 1.93/1.05 || 1.93/1.05 || quot(0, s(x0), s(x1)) 1.93/1.05 || quot(s(x0), s(x1), x2) 1.93/1.05 || quot(x0, 0, s(x1)) 1.93/1.05 || 1.93/1.05 || 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (8) 1.93/1.05 || Obligation: 1.93/1.05 || Q DP problem: 1.93/1.05 || The TRS P consists of the following rules: 1.93/1.05 || 1.93/1.05 || QUOT(s(%X), s(%Y), %Z) -> QUOT(%X, %Y, %Z) 1.93/1.05 || QUOT(%X, 0, s(%Y)) -> QUOT(%X, s(%Y), s(%Y)) 1.93/1.05 || 1.93/1.05 || R is empty. 1.93/1.05 || Q is empty. 1.93/1.05 || We have to consider all minimal (P,Q,R)-chains. 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (9) QDPSizeChangeProof (EQUIVALENT) 1.93/1.05 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 1.93/1.05 || 1.93/1.05 || From the DPs we obtained the following set of size-change graphs: 1.93/1.05 || *QUOT(s(%X), s(%Y), %Z) -> QUOT(%X, %Y, %Z) 1.93/1.05 || The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3 1.93/1.05 || 1.93/1.05 || 1.93/1.05 || *QUOT(%X, 0, s(%Y)) -> QUOT(%X, s(%Y), s(%Y)) 1.93/1.05 || The graph contains the following edges 1 >= 1, 3 >= 2, 3 >= 3 1.93/1.05 || 1.93/1.05 || 1.93/1.05 || ---------------------------------------- 1.93/1.05 || 1.93/1.05 || (10) 1.93/1.05 || YES 1.93/1.05 || 1.93/1.05 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 [Kop13]). 1.93/1.05 1.93/1.05 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 1.93/1.05 1.93/1.05 Dependency Pairs P_0: 1.93/1.05 1.93/1.05 0] map#(F, cons(X, Y)) =#> map#(F, Y) 1.93/1.05 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 1.93/1.05 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 1.93/1.05 3] filter2#(false, F, X, Y) =#> filter#(F, Y) 1.93/1.05 1.93/1.05 Rules R_0: 1.93/1.05 1.93/1.05 quot(0, s(X), s(Y)) => 0 1.93/1.05 quot(s(X), s(Y), Z) => quot(X, Y, Z) 1.93/1.05 quot(X, 0, s(Y)) => s(quot(X, s(Y), s(Y))) 1.93/1.05 map(F, nil) => nil 1.93/1.05 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 1.93/1.05 filter(F, nil) => nil 1.93/1.05 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 1.93/1.05 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 1.93/1.05 filter2(false, F, X, Y) => filter(F, Y) 1.93/1.05 1.93/1.05 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 1.93/1.05 1.93/1.05 We consider the dependency pair problem (P_0, R_0, static, formative). 1.93/1.05 1.93/1.05 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 1.93/1.05 1.93/1.05 * 0 : 0 1.93/1.05 * 1 : 2, 3 1.93/1.05 * 2 : 1 1.93/1.05 * 3 : 1 1.93/1.05 1.93/1.05 This graph has the following strongly connected components: 1.93/1.05 1.93/1.05 P_1: 1.93/1.05 1.93/1.05 map#(F, cons(X, Y)) =#> map#(F, Y) 1.93/1.05 1.93/1.05 P_2: 1.93/1.05 1.93/1.05 filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 1.93/1.05 filter2#(true, F, X, Y) =#> filter#(F, Y) 1.93/1.05 filter2#(false, F, X, Y) =#> filter#(F, Y) 1.93/1.05 1.93/1.05 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). 1.93/1.05 1.93/1.05 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite. 1.93/1.05 1.93/1.05 We consider the dependency pair problem (P_2, R_0, static, formative). 1.93/1.05 1.93/1.05 We apply the subterm criterion with the following projection function: 1.93/1.05 1.93/1.05 nu(filter2#) = 4 1.93/1.05 nu(filter#) = 2 1.93/1.05 1.93/1.05 Thus, we can orient the dependency pairs as follows: 1.93/1.05 1.93/1.05 nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 1.93/1.05 nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 1.93/1.05 nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 1.93/1.05 1.93/1.05 By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_2, R_0, static, f) by (P_3, R_0, static, f), where P_3 contains: 1.93/1.05 1.93/1.05 filter2#(true, F, X, Y) =#> filter#(F, Y) 1.93/1.05 filter2#(false, F, X, Y) =#> filter#(F, Y) 1.93/1.05 1.93/1.05 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite. 1.93/1.05 1.93/1.05 We consider the dependency pair problem (P_3, R_0, static, formative). 1.93/1.05 1.93/1.05 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 1.93/1.05 1.93/1.05 * 0 : 1.93/1.05 * 1 : 1.93/1.05 1.93/1.05 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 1.93/1.05 1.93/1.05 Thus, the original system is terminating if (P_1, R_0, static, formative) is finite. 1.93/1.05 1.93/1.05 We consider the dependency pair problem (P_1, R_0, static, formative). 1.93/1.05 1.93/1.05 We apply the subterm criterion with the following projection function: 1.93/1.05 1.93/1.05 nu(map#) = 2 1.93/1.05 1.93/1.05 Thus, we can orient the dependency pairs as follows: 1.93/1.05 1.93/1.05 nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 1.93/1.05 1.93/1.05 By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_1, R_0, static, f) by ({}, R_0, static, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. 1.93/1.05 1.93/1.05 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 1.93/1.05 1.93/1.05 1.93/1.05 +++ Citations +++ 1.93/1.05 1.93/1.05 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 1.93/1.05 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 1.93/1.05 [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. 1.93/1.05 EOF