/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: 0 : [] --> b cons : [c * d] --> d false : [] --> a filter : [c -> a * d] --> d filter2 : [a * c -> a * c * d] --> d if : [a * b * b] --> b le : [b * b] --> a map : [c -> c * d] --> d minus : [b * b] --> b nil : [] --> d p : [b] --> b s : [b] --> b true : [] --> a Rules: p(0) => 0 p(s(x)) => x le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) minus(x, y) => if(le(x, y), x, y) if(true, x, y) => 0 if(false, x, y) => s(minus(p(x), y)) 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: p(0) => 0 p(s(X)) => X le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(X, Y) => if(le(X, Y), X, Y) if(true, X, Y) => 0 if(false, X, Y) => s(minus(p(X), Y)) 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) Overlay + Local Confluence [EQUIVALENT] || (2) QTRS || (3) DependencyPairsProof [EQUIVALENT] || (4) QDP || (5) DependencyGraphProof [EQUIVALENT] || (6) AND || (7) QDP || (8) UsableRulesProof [EQUIVALENT] || (9) QDP || (10) QReductionProof [EQUIVALENT] || (11) QDP || (12) QDPSizeChangeProof [EQUIVALENT] || (13) YES || (14) QDP || (15) UsableRulesProof [EQUIVALENT] || (16) QDP || (17) QReductionProof [EQUIVALENT] || (18) QDP || (19) QDPOrderProof [EQUIVALENT] || (20) QDP || (21) DependencyGraphProof [EQUIVALENT] || (22) TRUE || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || p(0) -> 0 || p(s(%X)) -> %X || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || minus(%X, %Y) -> if(le(%X, %Y), %X, %Y) || if(true, %X, %Y) -> 0 || if(false, %X, %Y) -> s(minus(p(%X), %Y)) || || Q is empty. || || ---------------------------------------- || || (1) Overlay + Local Confluence (EQUIVALENT) || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || p(0) -> 0 || p(s(%X)) -> %X || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || minus(%X, %Y) -> if(le(%X, %Y), %X, %Y) || if(true, %X, %Y) -> 0 || if(false, %X, %Y) -> s(minus(p(%X), %Y)) || || The set Q consists of the following terms: || || p(0) || p(s(x0)) || le(0, x0) || le(s(x0), 0) || le(s(x0), s(x1)) || minus(x0, x1) || if(true, x0, x1) || if(false, x0, x1) || || || ---------------------------------------- || || (3) DependencyPairsProof (EQUIVALENT) || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. || ---------------------------------------- || || (4) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || LE(s(%X), s(%Y)) -> LE(%X, %Y) || MINUS(%X, %Y) -> IF(le(%X, %Y), %X, %Y) || MINUS(%X, %Y) -> LE(%X, %Y) || IF(false, %X, %Y) -> MINUS(p(%X), %Y) || IF(false, %X, %Y) -> P(%X) || || The TRS R consists of the following rules: || || p(0) -> 0 || p(s(%X)) -> %X || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || minus(%X, %Y) -> if(le(%X, %Y), %X, %Y) || if(true, %X, %Y) -> 0 || if(false, %X, %Y) -> s(minus(p(%X), %Y)) || || The set Q consists of the following terms: || || p(0) || p(s(x0)) || le(0, x0) || le(s(x0), 0) || le(s(x0), s(x1)) || minus(x0, x1) || if(true, x0, x1) || if(false, x0, x1) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (5) DependencyGraphProof (EQUIVALENT) || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes. || ---------------------------------------- || || (6) || Complex Obligation (AND) || || ---------------------------------------- || || (7) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || LE(s(%X), s(%Y)) -> LE(%X, %Y) || || The TRS R consists of the following rules: || || p(0) -> 0 || p(s(%X)) -> %X || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || minus(%X, %Y) -> if(le(%X, %Y), %X, %Y) || if(true, %X, %Y) -> 0 || if(false, %X, %Y) -> s(minus(p(%X), %Y)) || || The set Q consists of the following terms: || || p(0) || p(s(x0)) || le(0, x0) || le(s(x0), 0) || le(s(x0), s(x1)) || minus(x0, x1) || if(true, x0, x1) || if(false, x0, x1) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (8) UsableRulesProof (EQUIVALENT) || 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. || ---------------------------------------- || || (9) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || LE(s(%X), s(%Y)) -> LE(%X, %Y) || || R is empty. || The set Q consists of the following terms: || || p(0) || p(s(x0)) || le(0, x0) || le(s(x0), 0) || le(s(x0), s(x1)) || minus(x0, x1) || if(true, x0, x1) || if(false, x0, x1) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (10) QReductionProof (EQUIVALENT) || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. || || p(0) || p(s(x0)) || le(0, x0) || le(s(x0), 0) || le(s(x0), s(x1)) || minus(x0, x1) || if(true, x0, x1) || if(false, x0, x1) || || || ---------------------------------------- || || (11) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || LE(s(%X), s(%Y)) -> LE(%X, %Y) || || R is empty. || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (12) QDPSizeChangeProof (EQUIVALENT) || 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. || || From the DPs we obtained the following set of size-change graphs: || *LE(s(%X), s(%Y)) -> LE(%X, %Y) || The graph contains the following edges 1 > 1, 2 > 2 || || || ---------------------------------------- || || (13) || YES || || ---------------------------------------- || || (14) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(%X, %Y) -> IF(le(%X, %Y), %X, %Y) || IF(false, %X, %Y) -> MINUS(p(%X), %Y) || || The TRS R consists of the following rules: || || p(0) -> 0 || p(s(%X)) -> %X || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || minus(%X, %Y) -> if(le(%X, %Y), %X, %Y) || if(true, %X, %Y) -> 0 || if(false, %X, %Y) -> s(minus(p(%X), %Y)) || || The set Q consists of the following terms: || || p(0) || p(s(x0)) || le(0, x0) || le(s(x0), 0) || le(s(x0), s(x1)) || minus(x0, x1) || if(true, x0, x1) || if(false, x0, x1) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (15) UsableRulesProof (EQUIVALENT) || 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. || ---------------------------------------- || || (16) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(%X, %Y) -> IF(le(%X, %Y), %X, %Y) || IF(false, %X, %Y) -> MINUS(p(%X), %Y) || || The TRS R consists of the following rules: || || p(0) -> 0 || p(s(%X)) -> %X || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || || The set Q consists of the following terms: || || p(0) || p(s(x0)) || le(0, x0) || le(s(x0), 0) || le(s(x0), s(x1)) || minus(x0, x1) || if(true, x0, x1) || if(false, x0, x1) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (17) QReductionProof (EQUIVALENT) || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. || || minus(x0, x1) || if(true, x0, x1) || if(false, x0, x1) || || || ---------------------------------------- || || (18) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(%X, %Y) -> IF(le(%X, %Y), %X, %Y) || IF(false, %X, %Y) -> MINUS(p(%X), %Y) || || The TRS R consists of the following rules: || || p(0) -> 0 || p(s(%X)) -> %X || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || || The set Q consists of the following terms: || || p(0) || p(s(x0)) || le(0, x0) || le(s(x0), 0) || le(s(x0), s(x1)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (19) QDPOrderProof (EQUIVALENT) || We use the reduction pair processor [LPAR04,JAR06]. || || || The following pairs can be oriented strictly and are deleted. || || IF(false, %X, %Y) -> MINUS(p(%X), %Y) || The remaining pairs can at least be oriented weakly. || Used ordering: Polynomial interpretation [POLO,RATPOLO]: || || POL(0) = [1/4] || POL(IF(x_1, x_2, x_3)) = [1/2] + [1/4]x_1 + [1/4]x_2 + [1/4]x_3 || POL(MINUS(x_1, x_2)) = [1/2] + [1/2]x_1 + [1/4]x_2 || POL(false) = [1] || POL(le(x_1, x_2)) = [1/2]x_1 || POL(p(x_1)) = [1/4] + [1/2]x_1 || POL(s(x_1)) = [4] + [4]x_1 || POL(true) = 0 || The value of delta used in the strict ordering is 1/8. || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: || || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || p(0) -> 0 || p(s(%X)) -> %X || || || ---------------------------------------- || || (20) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(%X, %Y) -> IF(le(%X, %Y), %X, %Y) || || The TRS R consists of the following rules: || || p(0) -> 0 || p(s(%X)) -> %X || le(0, %X) -> true || le(s(%X), 0) -> false || le(s(%X), s(%Y)) -> le(%X, %Y) || || The set Q consists of the following terms: || || p(0) || p(s(x0)) || le(0, x0) || le(s(x0), 0) || le(s(x0), s(x1)) || || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (21) DependencyGraphProof (EQUIVALENT) || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. || ---------------------------------------- || || (22) || TRUE || We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, 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: p(0) => 0 p(s(X)) => X le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(X, Y) => if(le(X, Y), X, Y) if(true, X, Y) => 0 if(false, X, Y) => s(minus(p(X), Y)) 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, 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 : 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, 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(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 [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_3, R_0, minimal, 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, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, 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 : 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, 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(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 [Kop12, Thm. 7.35], 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 +++ [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.