/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 : [] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d minus : [a * a] --> a nil : [] --> d plus : [a * a] --> a quot : [a * a] --> a s : [a] --> a true : [] --> b Rules: minus(x, 0) => x minus(s(x), s(y)) => minus(x, y) quot(0, s(x)) => 0 quot(s(x), s(y)) => s(quot(minus(x, y), s(y))) plus(0, x) => x plus(s(x), y) => s(plus(x, y)) plus(minus(x, s(0)), minus(y, s(s(z)))) => plus(minus(y, s(s(z))), minus(x, s(0))) plus(plus(x, s(0)), plus(y, s(s(z)))) => plus(plus(y, s(s(z))), plus(x, s(0))) 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: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) plus(minus(X, s(0)), minus(Y, s(s(Z)))) => plus(minus(Y, s(s(Z))), minus(X, s(0))) plus(plus(X, s(0)), plus(Y, s(s(Z)))) => plus(plus(Y, s(s(Z))), plus(X, s(0))) Moreover, the system is finitely branching. 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 Ce-terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed Ce-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) DependencyPairsProof [EQUIVALENT] || (2) QDP || (3) DependencyGraphProof [EQUIVALENT] || (4) AND || (5) QDP || (6) UsableRulesProof [EQUIVALENT] || (7) QDP || (8) MRRProof [EQUIVALENT] || (9) QDP || (10) MRRProof [EQUIVALENT] || (11) QDP || (12) MRRProof [EQUIVALENT] || (13) QDP || (14) DependencyGraphProof [EQUIVALENT] || (15) TRUE || (16) QDP || (17) UsableRulesProof [EQUIVALENT] || (18) QDP || (19) QDPSizeChangeProof [EQUIVALENT] || (20) YES || (21) QDP || (22) QDPOrderProof [EQUIVALENT] || (23) QDP || (24) PisEmptyProof [EQUIVALENT] || (25) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(%X, %Y) || quot(0, s(%X)) -> 0 || quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y))) || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0))) || plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0))) || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || || ---------------------------------------- || || (1) DependencyPairsProof (EQUIVALENT) || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. || ---------------------------------------- || || (2) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y) || QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y)) || QUOT(s(%X), s(%Y)) -> MINUS(%X, %Y) || PLUS(s(%X), %Y) -> PLUS(%X, %Y) || PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0))) || PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0))) || || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(%X, %Y) || quot(0, s(%X)) -> 0 || quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y))) || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0))) || plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0))) || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (3) DependencyGraphProof (EQUIVALENT) || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 1 less node. || ---------------------------------------- || || (4) || Complex Obligation (AND) || || ---------------------------------------- || || (5) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0))) || PLUS(s(%X), %Y) -> PLUS(%X, %Y) || PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0))) || || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(%X, %Y) || quot(0, s(%X)) -> 0 || quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y))) || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0))) || plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0))) || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (6) UsableRulesProof (EQUIVALENT) || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. || ---------------------------------------- || || (7) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0))) || PLUS(s(%X), %Y) -> PLUS(%X, %Y) || PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0))) || || The TRS R consists of the following rules: || || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0))) || plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0))) || minus(s(%X), s(%Y)) -> minus(%X, %Y) || minus(%X, 0) -> %X || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (8) MRRProof (EQUIVALENT) || By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. || || || Strictly oriented rules of the TRS R: || || minus(%X, 0) -> %X || || Used ordering: Polynomial interpretation [POLO]: || || POL(0) = 0 || POL(PLUS(x_1, x_2)) = 2*x_1 + 2*x_2 || POL(minus(x_1, x_2)) = 2 + x_1 + x_2 || POL(plus(x_1, x_2)) = x_1 + x_2 || POL(s(x_1)) = x_1 || || || ---------------------------------------- || || (9) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0))) || PLUS(s(%X), %Y) -> PLUS(%X, %Y) || PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0))) || || The TRS R consists of the following rules: || || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0))) || plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0))) || minus(s(%X), s(%Y)) -> minus(%X, %Y) || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (10) MRRProof (EQUIVALENT) || By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. || || || Strictly oriented rules of the TRS R: || || plus(0, %X) -> %X || || Used ordering: Polynomial interpretation [POLO]: || || POL(0) = 0 || POL(PLUS(x_1, x_2)) = x_1 + x_2 || POL(minus(x_1, x_2)) = 2*x_1 + 2*x_2 || POL(plus(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 || POL(s(x_1)) = x_1 || || || ---------------------------------------- || || (11) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0))) || PLUS(s(%X), %Y) -> PLUS(%X, %Y) || PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0))) || || The TRS R consists of the following rules: || || plus(s(%X), %Y) -> s(plus(%X, %Y)) || plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0))) || plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0))) || minus(s(%X), s(%Y)) -> minus(%X, %Y) || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (12) MRRProof (EQUIVALENT) || By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. || || Strictly oriented dependency pairs: || || PLUS(s(%X), %Y) -> PLUS(%X, %Y) || || Strictly oriented rules of the TRS R: || || minus(s(%X), s(%Y)) -> minus(%X, %Y) || || Used ordering: Polynomial interpretation [POLO]: || || POL(0) = 0 || POL(PLUS(x_1, x_2)) = 2*x_1 + 2*x_2 || POL(minus(x_1, x_2)) = 2*x_1 + 2*x_2 || POL(plus(x_1, x_2)) = x_1 + x_2 || POL(s(x_1)) = 1 + x_1 || || || ---------------------------------------- || || (13) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0))) || PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0))) || || The TRS R consists of the following rules: || || plus(s(%X), %Y) -> s(plus(%X, %Y)) || plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0))) || plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0))) || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (14) DependencyGraphProof (EQUIVALENT) || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes. || ---------------------------------------- || || (15) || TRUE || || ---------------------------------------- || || (16) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y) || || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(%X, %Y) || quot(0, s(%X)) -> 0 || quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y))) || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0))) || plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0))) || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (17) UsableRulesProof (EQUIVALENT) || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. || ---------------------------------------- || || (18) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y) || || R is empty. || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (19) 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: || *MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y) || The graph contains the following edges 1 > 1, 2 > 2 || || || ---------------------------------------- || || (20) || YES || || ---------------------------------------- || || (21) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y)) || || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(%X, %Y) || quot(0, s(%X)) -> 0 || quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y))) || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0))) || plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0))) || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (22) QDPOrderProof (EQUIVALENT) || We use the reduction pair processor [LPAR04,JAR06]. || || || The following pairs can be oriented strictly and are deleted. || || QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y)) || The remaining pairs can at least be oriented weakly. || Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: || || POL( QUOT_2(x_1, x_2) ) = 2x_1 + x_2 || POL( minus_2(x_1, x_2) ) = x_1 + 1 || POL( 0 ) = 2 || POL( s_1(x_1) ) = 2x_1 + 2 || || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(%X, %Y) || || || ---------------------------------------- || || (23) || Obligation: || Q DP problem: || P is empty. || The TRS R consists of the following rules: || || minus(%X, 0) -> %X || minus(s(%X), s(%Y)) -> minus(%X, %Y) || quot(0, s(%X)) -> 0 || quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y))) || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0))) || plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0))) || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (24) PisEmptyProof (EQUIVALENT) || The TRS P is empty. Hence, there is no (P,Q,R) chain. || ---------------------------------------- || || (25) || 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 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: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) plus(minus(X, s(0)), minus(Y, s(s(Z)))) => plus(minus(Y, s(s(Z))), minus(X, s(0))) plus(plus(X, s(0)), plus(Y, s(s(Z)))) => plus(plus(Y, s(s(Z))), plus(X, s(0))) 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.