2.17/1.17 YES 2.17/1.18 We consider the system theBenchmark. 2.17/1.18 2.17/1.18 Alphabet: 2.17/1.18 2.17/1.18 0 : [] --> b 2.17/1.18 cons : [d * e] --> e 2.17/1.18 f : [b * b * b * b] --> c 2.17/1.18 false : [] --> c 2.17/1.18 filter : [d -> c * e] --> e 2.17/1.18 filter2 : [c * d -> c * d * e] --> e 2.17/1.18 if : [a * c * c] --> c 2.17/1.18 le : [b * b] --> a 2.17/1.18 map : [d -> d * e] --> e 2.17/1.18 minus : [b * b] --> b 2.17/1.18 nil : [] --> e 2.17/1.18 perfectp : [b] --> c 2.17/1.18 s : [b] --> b 2.17/1.18 true : [] --> c 2.17/1.18 2.17/1.18 Rules: 2.17/1.18 2.17/1.18 perfectp(0) => false 2.17/1.18 perfectp(s(x)) => f(x, s(0), s(x), s(x)) 2.17/1.18 f(0, x, 0, y) => true 2.17/1.18 f(0, x, s(y), z) => false 2.17/1.18 f(s(x), 0, y, z) => f(x, z, minus(y, s(x)), z) 2.17/1.18 f(s(x), s(y), z, u) => if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) 2.17/1.18 map(g, nil) => nil 2.17/1.18 map(g, cons(x, y)) => cons(g x, map(g, y)) 2.17/1.18 filter(g, nil) => nil 2.17/1.18 filter(g, cons(x, y)) => filter2(g x, g, x, y) 2.17/1.18 filter2(true, g, x, y) => cons(x, filter(g, y)) 2.17/1.18 filter2(false, g, x, y) => filter(g, y) 2.17/1.18 2.17/1.18 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 2.17/1.18 2.17/1.18 We observe that the rules contain a first-order subset: 2.17/1.18 2.17/1.18 perfectp(0) => false 2.17/1.18 perfectp(s(X)) => f(X, s(0), s(X), s(X)) 2.17/1.18 f(0, X, 0, Y) => true 2.17/1.18 f(0, X, s(Y), Z) => false 2.17/1.18 f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 2.17/1.18 f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 2.17/1.18 2.17/1.18 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. 2.17/1.18 2.17/1.18 According to the external first-order termination prover, this system is indeed terminating: 2.17/1.18 2.17/1.18 || proof of resources/system.trs 2.17/1.18 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 2.17/1.18 || 2.17/1.18 || 2.17/1.18 || Termination w.r.t. Q of the given QTRS could be proven: 2.17/1.18 || 2.17/1.18 || (0) QTRS 2.17/1.18 || (1) Overlay + Local Confluence [EQUIVALENT] 2.17/1.18 || (2) QTRS 2.17/1.18 || (3) DependencyPairsProof [EQUIVALENT] 2.17/1.18 || (4) QDP 2.17/1.18 || (5) DependencyGraphProof [EQUIVALENT] 2.17/1.18 || (6) QDP 2.17/1.18 || (7) UsableRulesProof [EQUIVALENT] 2.17/1.18 || (8) QDP 2.17/1.18 || (9) QReductionProof [EQUIVALENT] 2.17/1.18 || (10) QDP 2.17/1.18 || (11) QDPSizeChangeProof [EQUIVALENT] 2.17/1.18 || (12) YES 2.17/1.18 || 2.17/1.18 || 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (0) 2.17/1.18 || Obligation: 2.17/1.18 || Q restricted rewrite system: 2.17/1.18 || The TRS R consists of the following rules: 2.17/1.18 || 2.17/1.18 || perfectp(0) -> false 2.17/1.18 || perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X)) 2.17/1.18 || f(0, %X, 0, %Y) -> true 2.17/1.18 || f(0, %X, s(%Y), %Z) -> false 2.17/1.18 || f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z) 2.17/1.18 || f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U)) 2.17/1.18 || 2.17/1.18 || Q is empty. 2.17/1.18 || 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (1) Overlay + Local Confluence (EQUIVALENT) 2.17/1.18 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (2) 2.17/1.18 || Obligation: 2.17/1.18 || Q restricted rewrite system: 2.17/1.18 || The TRS R consists of the following rules: 2.17/1.18 || 2.17/1.18 || perfectp(0) -> false 2.17/1.18 || perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X)) 2.17/1.18 || f(0, %X, 0, %Y) -> true 2.17/1.18 || f(0, %X, s(%Y), %Z) -> false 2.17/1.18 || f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z) 2.17/1.18 || f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U)) 2.17/1.18 || 2.17/1.18 || The set Q consists of the following terms: 2.17/1.18 || 2.17/1.18 || perfectp(0) 2.17/1.18 || perfectp(s(x0)) 2.17/1.18 || f(0, x0, 0, x1) 2.17/1.18 || f(0, x0, s(x1), x2) 2.17/1.18 || f(s(x0), 0, x1, x2) 2.17/1.18 || f(s(x0), s(x1), x2, x3) 2.17/1.18 || 2.17/1.18 || 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (3) DependencyPairsProof (EQUIVALENT) 2.17/1.18 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (4) 2.17/1.18 || Obligation: 2.17/1.18 || Q DP problem: 2.17/1.18 || The TRS P consists of the following rules: 2.17/1.18 || 2.17/1.18 || PERFECTP(s(%X)) -> F(%X, s(0), s(%X), s(%X)) 2.17/1.18 || F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z) 2.17/1.18 || F(s(%X), s(%Y), %Z, %U) -> F(s(%X), minus(%Y, %X), %Z, %U) 2.17/1.18 || F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U) 2.17/1.18 || 2.17/1.18 || The TRS R consists of the following rules: 2.17/1.18 || 2.17/1.18 || perfectp(0) -> false 2.17/1.18 || perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X)) 2.17/1.18 || f(0, %X, 0, %Y) -> true 2.17/1.18 || f(0, %X, s(%Y), %Z) -> false 2.17/1.18 || f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z) 2.17/1.18 || f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U)) 2.17/1.18 || 2.17/1.18 || The set Q consists of the following terms: 2.17/1.18 || 2.17/1.18 || perfectp(0) 2.17/1.18 || perfectp(s(x0)) 2.17/1.18 || f(0, x0, 0, x1) 2.17/1.18 || f(0, x0, s(x1), x2) 2.17/1.18 || f(s(x0), 0, x1, x2) 2.17/1.18 || f(s(x0), s(x1), x2, x3) 2.17/1.18 || 2.17/1.18 || We have to consider all minimal (P,Q,R)-chains. 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (5) DependencyGraphProof (EQUIVALENT) 2.17/1.18 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (6) 2.17/1.18 || Obligation: 2.17/1.18 || Q DP problem: 2.17/1.18 || The TRS P consists of the following rules: 2.17/1.18 || 2.17/1.18 || F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U) 2.17/1.18 || F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z) 2.17/1.18 || 2.17/1.18 || The TRS R consists of the following rules: 2.17/1.18 || 2.17/1.18 || perfectp(0) -> false 2.17/1.18 || perfectp(s(%X)) -> f(%X, s(0), s(%X), s(%X)) 2.17/1.18 || f(0, %X, 0, %Y) -> true 2.17/1.18 || f(0, %X, s(%Y), %Z) -> false 2.17/1.18 || f(s(%X), 0, %Y, %Z) -> f(%X, %Z, minus(%Y, s(%X)), %Z) 2.17/1.18 || f(s(%X), s(%Y), %Z, %U) -> if(le(%X, %Y), f(s(%X), minus(%Y, %X), %Z, %U), f(%X, %U, %Z, %U)) 2.17/1.18 || 2.17/1.18 || The set Q consists of the following terms: 2.17/1.18 || 2.17/1.18 || perfectp(0) 2.17/1.18 || perfectp(s(x0)) 2.17/1.18 || f(0, x0, 0, x1) 2.17/1.18 || f(0, x0, s(x1), x2) 2.17/1.18 || f(s(x0), 0, x1, x2) 2.17/1.18 || f(s(x0), s(x1), x2, x3) 2.17/1.18 || 2.17/1.18 || We have to consider all minimal (P,Q,R)-chains. 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (7) UsableRulesProof (EQUIVALENT) 2.17/1.18 || 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. 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (8) 2.17/1.18 || Obligation: 2.17/1.18 || Q DP problem: 2.17/1.18 || The TRS P consists of the following rules: 2.17/1.18 || 2.17/1.18 || F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U) 2.17/1.18 || F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z) 2.17/1.18 || 2.17/1.18 || R is empty. 2.17/1.18 || The set Q consists of the following terms: 2.17/1.18 || 2.17/1.18 || perfectp(0) 2.17/1.18 || perfectp(s(x0)) 2.17/1.18 || f(0, x0, 0, x1) 2.17/1.18 || f(0, x0, s(x1), x2) 2.17/1.18 || f(s(x0), 0, x1, x2) 2.17/1.18 || f(s(x0), s(x1), x2, x3) 2.17/1.18 || 2.17/1.18 || We have to consider all minimal (P,Q,R)-chains. 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (9) QReductionProof (EQUIVALENT) 2.17/1.18 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 2.17/1.18 || 2.17/1.18 || perfectp(0) 2.17/1.18 || perfectp(s(x0)) 2.17/1.18 || f(0, x0, 0, x1) 2.17/1.18 || f(0, x0, s(x1), x2) 2.17/1.18 || f(s(x0), 0, x1, x2) 2.17/1.18 || f(s(x0), s(x1), x2, x3) 2.17/1.18 || 2.17/1.18 || 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (10) 2.17/1.18 || Obligation: 2.17/1.18 || Q DP problem: 2.17/1.18 || The TRS P consists of the following rules: 2.17/1.18 || 2.17/1.18 || F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U) 2.17/1.18 || F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z) 2.17/1.18 || 2.17/1.18 || R is empty. 2.17/1.18 || Q is empty. 2.17/1.18 || We have to consider all minimal (P,Q,R)-chains. 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (11) QDPSizeChangeProof (EQUIVALENT) 2.17/1.18 || 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. 2.17/1.18 || 2.17/1.18 || From the DPs we obtained the following set of size-change graphs: 2.17/1.18 || *F(s(%X), s(%Y), %Z, %U) -> F(%X, %U, %Z, %U) 2.17/1.18 || The graph contains the following edges 1 > 1, 4 >= 2, 3 >= 3, 4 >= 4 2.17/1.18 || 2.17/1.18 || 2.17/1.18 || *F(s(%X), 0, %Y, %Z) -> F(%X, %Z, minus(%Y, s(%X)), %Z) 2.17/1.18 || The graph contains the following edges 1 > 1, 4 >= 2, 4 >= 4 2.17/1.18 || 2.17/1.18 || 2.17/1.18 || ---------------------------------------- 2.17/1.18 || 2.17/1.18 || (12) 2.17/1.18 || YES 2.17/1.18 || 2.17/1.18 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]). 2.17/1.18 2.17/1.18 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 2.17/1.18 2.17/1.18 Dependency Pairs P_0: 2.17/1.18 2.17/1.18 0] map#(F, cons(X, Y)) =#> map#(F, Y) 2.17/1.18 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2.17/1.18 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 2.17/1.18 3] filter2#(false, F, X, Y) =#> filter#(F, Y) 2.17/1.18 2.17/1.18 Rules R_0: 2.17/1.18 2.17/1.18 perfectp(0) => false 2.17/1.18 perfectp(s(X)) => f(X, s(0), s(X), s(X)) 2.17/1.18 f(0, X, 0, Y) => true 2.17/1.18 f(0, X, s(Y), Z) => false 2.17/1.18 f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) 2.17/1.18 f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 2.17/1.18 map(F, nil) => nil 2.17/1.18 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 2.17/1.18 filter(F, nil) => nil 2.17/1.18 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 2.17/1.18 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 2.17/1.18 filter2(false, F, X, Y) => filter(F, Y) 2.17/1.18 2.17/1.18 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 2.17/1.18 2.17/1.18 We consider the dependency pair problem (P_0, R_0, static, formative). 2.17/1.18 2.17/1.18 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 2.17/1.18 2.17/1.18 * 0 : 0 2.17/1.18 * 1 : 2, 3 2.17/1.18 * 2 : 1 2.17/1.18 * 3 : 1 2.17/1.18 2.17/1.18 This graph has the following strongly connected components: 2.17/1.18 2.17/1.18 P_1: 2.17/1.18 2.17/1.18 map#(F, cons(X, Y)) =#> map#(F, Y) 2.17/1.18 2.17/1.18 P_2: 2.17/1.18 2.17/1.18 filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2.17/1.18 filter2#(true, F, X, Y) =#> filter#(F, Y) 2.17/1.18 filter2#(false, F, X, Y) =#> filter#(F, Y) 2.17/1.18 2.17/1.18 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). 2.17/1.18 2.17/1.18 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite. 2.17/1.18 2.17/1.18 We consider the dependency pair problem (P_2, R_0, static, formative). 2.17/1.18 2.17/1.18 We apply the subterm criterion with the following projection function: 2.17/1.18 2.17/1.18 nu(filter2#) = 4 2.17/1.18 nu(filter#) = 2 2.17/1.18 2.17/1.18 Thus, we can orient the dependency pairs as follows: 2.17/1.18 2.17/1.18 nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 2.17/1.18 nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 2.17/1.18 nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 2.17/1.18 2.17/1.18 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: 2.17/1.18 2.17/1.18 filter2#(true, F, X, Y) =#> filter#(F, Y) 2.17/1.18 filter2#(false, F, X, Y) =#> filter#(F, Y) 2.17/1.18 2.17/1.18 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite. 2.17/1.18 2.17/1.18 We consider the dependency pair problem (P_3, R_0, static, formative). 2.17/1.18 2.17/1.18 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 2.17/1.18 2.17/1.18 * 0 : 2.17/1.18 * 1 : 2.17/1.18 2.17/1.18 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 2.17/1.18 2.17/1.18 Thus, the original system is terminating if (P_1, R_0, static, formative) is finite. 2.17/1.18 2.17/1.18 We consider the dependency pair problem (P_1, R_0, static, formative). 2.17/1.18 2.17/1.18 We apply the subterm criterion with the following projection function: 2.17/1.18 2.17/1.18 nu(map#) = 2 2.17/1.18 2.17/1.18 Thus, we can orient the dependency pairs as follows: 2.17/1.18 2.17/1.18 nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 2.17/1.18 2.17/1.18 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. 2.17/1.18 2.17/1.18 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 2.17/1.18 2.17/1.18 2.17/1.18 +++ Citations +++ 2.17/1.18 2.17/1.18 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 2.17/1.18 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 2.17/1.18 [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. 2.17/1.18 EOF