2.57/1.26 YES 2.59/1.27 We consider the system theBenchmark. 2.59/1.27 2.59/1.27 Alphabet: 2.59/1.27 2.59/1.27 0 : [] --> b 2.59/1.27 cons : [b * c] --> c 2.59/1.27 false : [] --> a 2.59/1.27 filter : [b -> a * c] --> c 2.59/1.27 filter2 : [a * b -> a * b * c] --> c 2.59/1.27 int : [b * b] --> c 2.59/1.27 intlist : [c] --> c 2.59/1.27 map : [b -> b * c] --> c 2.59/1.27 nil : [] --> c 2.59/1.27 s : [b] --> b 2.59/1.27 true : [] --> a 2.59/1.27 2.59/1.27 Rules: 2.59/1.27 2.59/1.27 intlist(nil) => nil 2.59/1.27 intlist(cons(x, y)) => cons(s(x), intlist(y)) 2.59/1.27 int(0, 0) => cons(0, nil) 2.59/1.27 int(0, s(x)) => cons(0, int(s(0), s(x))) 2.59/1.27 int(s(x), 0) => nil 2.59/1.27 int(s(x), s(y)) => intlist(int(x, y)) 2.59/1.27 map(f, nil) => nil 2.59/1.27 map(f, cons(x, y)) => cons(f x, map(f, y)) 2.59/1.27 filter(f, nil) => nil 2.59/1.27 filter(f, cons(x, y)) => filter2(f x, f, x, y) 2.59/1.27 filter2(true, f, x, y) => cons(x, filter(f, y)) 2.59/1.27 filter2(false, f, x, y) => filter(f, y) 2.59/1.27 2.59/1.27 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 2.59/1.27 2.59/1.27 We observe that the rules contain a first-order subset: 2.59/1.27 2.59/1.27 intlist(nil) => nil 2.59/1.27 intlist(cons(X, Y)) => cons(s(X), intlist(Y)) 2.59/1.27 int(0, 0) => cons(0, nil) 2.59/1.27 int(0, s(X)) => cons(0, int(s(0), s(X))) 2.59/1.27 int(s(X), 0) => nil 2.59/1.27 int(s(X), s(Y)) => intlist(int(X, Y)) 2.59/1.27 2.59/1.27 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. 2.59/1.27 2.59/1.27 According to the external first-order termination prover, this system is indeed terminating: 2.59/1.27 2.59/1.27 || proof of resources/system.trs 2.59/1.27 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || Termination w.r.t. Q of the given QTRS could be proven: 2.59/1.27 || 2.59/1.27 || (0) QTRS 2.59/1.27 || (1) QTRSRRRProof [EQUIVALENT] 2.59/1.27 || (2) QTRS 2.59/1.27 || (3) QTRSRRRProof [EQUIVALENT] 2.59/1.27 || (4) QTRS 2.59/1.27 || (5) Overlay + Local Confluence [EQUIVALENT] 2.59/1.27 || (6) QTRS 2.59/1.27 || (7) DependencyPairsProof [EQUIVALENT] 2.59/1.27 || (8) QDP 2.59/1.27 || (9) DependencyGraphProof [EQUIVALENT] 2.59/1.27 || (10) AND 2.59/1.27 || (11) QDP 2.59/1.27 || (12) UsableRulesProof [EQUIVALENT] 2.59/1.27 || (13) QDP 2.59/1.27 || (14) QReductionProof [EQUIVALENT] 2.59/1.27 || (15) QDP 2.59/1.27 || (16) QDPSizeChangeProof [EQUIVALENT] 2.59/1.27 || (17) YES 2.59/1.27 || (18) QDP 2.59/1.27 || (19) UsableRulesProof [EQUIVALENT] 2.59/1.27 || (20) QDP 2.59/1.27 || (21) QReductionProof [EQUIVALENT] 2.59/1.27 || (22) QDP 2.59/1.27 || (23) QDPSizeChangeProof [EQUIVALENT] 2.59/1.27 || (24) YES 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (0) 2.59/1.27 || Obligation: 2.59/1.27 || Q restricted rewrite system: 2.59/1.27 || The TRS R consists of the following rules: 2.59/1.27 || 2.59/1.27 || intlist(nil) -> nil 2.59/1.27 || intlist(cons(%X, %Y)) -> cons(s(%X), intlist(%Y)) 2.59/1.27 || int(0, 0) -> cons(0, nil) 2.59/1.27 || int(0, s(%X)) -> cons(0, int(s(0), s(%X))) 2.59/1.27 || int(s(%X), 0) -> nil 2.59/1.27 || int(s(%X), s(%Y)) -> intlist(int(%X, %Y)) 2.59/1.27 || 2.59/1.27 || Q is empty. 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (1) QTRSRRRProof (EQUIVALENT) 2.59/1.27 || Used ordering: 2.59/1.27 || Polynomial interpretation [POLO]: 2.59/1.27 || 2.59/1.27 || POL(0) = 0 2.59/1.27 || POL(cons(x_1, x_2)) = 2*x_1 + x_2 2.59/1.27 || POL(int(x_1, x_2)) = 1 + 2*x_1 + x_2 2.59/1.27 || POL(intlist(x_1)) = x_1 2.59/1.27 || POL(nil) = 0 2.59/1.27 || POL(s(x_1)) = x_1 2.59/1.27 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 2.59/1.27 || 2.59/1.27 || int(0, 0) -> cons(0, nil) 2.59/1.27 || int(s(%X), 0) -> nil 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (2) 2.59/1.27 || Obligation: 2.59/1.27 || Q restricted rewrite system: 2.59/1.27 || The TRS R consists of the following rules: 2.59/1.27 || 2.59/1.27 || intlist(nil) -> nil 2.59/1.27 || intlist(cons(%X, %Y)) -> cons(s(%X), intlist(%Y)) 2.59/1.27 || int(0, s(%X)) -> cons(0, int(s(0), s(%X))) 2.59/1.27 || int(s(%X), s(%Y)) -> intlist(int(%X, %Y)) 2.59/1.27 || 2.59/1.27 || Q is empty. 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (3) QTRSRRRProof (EQUIVALENT) 2.59/1.27 || Used ordering: 2.59/1.27 || Polynomial interpretation [POLO]: 2.59/1.27 || 2.59/1.27 || POL(0) = 0 2.59/1.27 || POL(cons(x_1, x_2)) = 2*x_1 + x_2 2.59/1.27 || POL(int(x_1, x_2)) = 2*x_1 + x_2 2.59/1.27 || POL(intlist(x_1)) = 2*x_1 2.59/1.27 || POL(nil) = 2 2.59/1.27 || POL(s(x_1)) = 2*x_1 2.59/1.27 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 2.59/1.27 || 2.59/1.27 || intlist(nil) -> nil 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (4) 2.59/1.27 || Obligation: 2.59/1.27 || Q restricted rewrite system: 2.59/1.27 || The TRS R consists of the following rules: 2.59/1.27 || 2.59/1.27 || intlist(cons(%X, %Y)) -> cons(s(%X), intlist(%Y)) 2.59/1.27 || int(0, s(%X)) -> cons(0, int(s(0), s(%X))) 2.59/1.27 || int(s(%X), s(%Y)) -> intlist(int(%X, %Y)) 2.59/1.27 || 2.59/1.27 || Q is empty. 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (5) Overlay + Local Confluence (EQUIVALENT) 2.59/1.27 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (6) 2.59/1.27 || Obligation: 2.59/1.27 || Q restricted rewrite system: 2.59/1.27 || The TRS R consists of the following rules: 2.59/1.27 || 2.59/1.27 || intlist(cons(%X, %Y)) -> cons(s(%X), intlist(%Y)) 2.59/1.27 || int(0, s(%X)) -> cons(0, int(s(0), s(%X))) 2.59/1.27 || int(s(%X), s(%Y)) -> intlist(int(%X, %Y)) 2.59/1.27 || 2.59/1.27 || The set Q consists of the following terms: 2.59/1.27 || 2.59/1.27 || intlist(cons(x0, x1)) 2.59/1.27 || int(0, s(x0)) 2.59/1.27 || int(s(x0), s(x1)) 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (7) DependencyPairsProof (EQUIVALENT) 2.59/1.27 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (8) 2.59/1.27 || Obligation: 2.59/1.27 || Q DP problem: 2.59/1.27 || The TRS P consists of the following rules: 2.59/1.27 || 2.59/1.27 || INTLIST(cons(%X, %Y)) -> INTLIST(%Y) 2.59/1.27 || INT(0, s(%X)) -> INT(s(0), s(%X)) 2.59/1.27 || INT(s(%X), s(%Y)) -> INTLIST(int(%X, %Y)) 2.59/1.27 || INT(s(%X), s(%Y)) -> INT(%X, %Y) 2.59/1.27 || 2.59/1.27 || The TRS R consists of the following rules: 2.59/1.27 || 2.59/1.27 || intlist(cons(%X, %Y)) -> cons(s(%X), intlist(%Y)) 2.59/1.27 || int(0, s(%X)) -> cons(0, int(s(0), s(%X))) 2.59/1.27 || int(s(%X), s(%Y)) -> intlist(int(%X, %Y)) 2.59/1.27 || 2.59/1.27 || The set Q consists of the following terms: 2.59/1.27 || 2.59/1.27 || intlist(cons(x0, x1)) 2.59/1.27 || int(0, s(x0)) 2.59/1.27 || int(s(x0), s(x1)) 2.59/1.27 || 2.59/1.27 || We have to consider all minimal (P,Q,R)-chains. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (9) DependencyGraphProof (EQUIVALENT) 2.59/1.27 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (10) 2.59/1.27 || Complex Obligation (AND) 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (11) 2.59/1.27 || Obligation: 2.59/1.27 || Q DP problem: 2.59/1.27 || The TRS P consists of the following rules: 2.59/1.27 || 2.59/1.27 || INTLIST(cons(%X, %Y)) -> INTLIST(%Y) 2.59/1.27 || 2.59/1.27 || The TRS R consists of the following rules: 2.59/1.27 || 2.59/1.27 || intlist(cons(%X, %Y)) -> cons(s(%X), intlist(%Y)) 2.59/1.27 || int(0, s(%X)) -> cons(0, int(s(0), s(%X))) 2.59/1.27 || int(s(%X), s(%Y)) -> intlist(int(%X, %Y)) 2.59/1.27 || 2.59/1.27 || The set Q consists of the following terms: 2.59/1.27 || 2.59/1.27 || intlist(cons(x0, x1)) 2.59/1.27 || int(0, s(x0)) 2.59/1.27 || int(s(x0), s(x1)) 2.59/1.27 || 2.59/1.27 || We have to consider all minimal (P,Q,R)-chains. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (12) UsableRulesProof (EQUIVALENT) 2.59/1.27 || 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. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (13) 2.59/1.27 || Obligation: 2.59/1.27 || Q DP problem: 2.59/1.27 || The TRS P consists of the following rules: 2.59/1.27 || 2.59/1.27 || INTLIST(cons(%X, %Y)) -> INTLIST(%Y) 2.59/1.27 || 2.59/1.27 || R is empty. 2.59/1.27 || The set Q consists of the following terms: 2.59/1.27 || 2.59/1.27 || intlist(cons(x0, x1)) 2.59/1.27 || int(0, s(x0)) 2.59/1.27 || int(s(x0), s(x1)) 2.59/1.27 || 2.59/1.27 || We have to consider all minimal (P,Q,R)-chains. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (14) QReductionProof (EQUIVALENT) 2.59/1.27 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 2.59/1.27 || 2.59/1.27 || intlist(cons(x0, x1)) 2.59/1.27 || int(0, s(x0)) 2.59/1.27 || int(s(x0), s(x1)) 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (15) 2.59/1.27 || Obligation: 2.59/1.27 || Q DP problem: 2.59/1.27 || The TRS P consists of the following rules: 2.59/1.27 || 2.59/1.27 || INTLIST(cons(%X, %Y)) -> INTLIST(%Y) 2.59/1.27 || 2.59/1.27 || R is empty. 2.59/1.27 || Q is empty. 2.59/1.27 || We have to consider all minimal (P,Q,R)-chains. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (16) QDPSizeChangeProof (EQUIVALENT) 2.59/1.27 || 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. 2.59/1.27 || 2.59/1.27 || From the DPs we obtained the following set of size-change graphs: 2.59/1.27 || *INTLIST(cons(%X, %Y)) -> INTLIST(%Y) 2.59/1.27 || The graph contains the following edges 1 > 1 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (17) 2.59/1.27 || YES 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (18) 2.59/1.27 || Obligation: 2.59/1.27 || Q DP problem: 2.59/1.27 || The TRS P consists of the following rules: 2.59/1.27 || 2.59/1.27 || INT(s(%X), s(%Y)) -> INT(%X, %Y) 2.59/1.27 || INT(0, s(%X)) -> INT(s(0), s(%X)) 2.59/1.27 || 2.59/1.27 || The TRS R consists of the following rules: 2.59/1.27 || 2.59/1.27 || intlist(cons(%X, %Y)) -> cons(s(%X), intlist(%Y)) 2.59/1.27 || int(0, s(%X)) -> cons(0, int(s(0), s(%X))) 2.59/1.27 || int(s(%X), s(%Y)) -> intlist(int(%X, %Y)) 2.59/1.27 || 2.59/1.27 || The set Q consists of the following terms: 2.59/1.27 || 2.59/1.27 || intlist(cons(x0, x1)) 2.59/1.27 || int(0, s(x0)) 2.59/1.27 || int(s(x0), s(x1)) 2.59/1.27 || 2.59/1.27 || We have to consider all minimal (P,Q,R)-chains. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (19) UsableRulesProof (EQUIVALENT) 2.59/1.27 || 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. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (20) 2.59/1.27 || Obligation: 2.59/1.27 || Q DP problem: 2.59/1.27 || The TRS P consists of the following rules: 2.59/1.27 || 2.59/1.27 || INT(s(%X), s(%Y)) -> INT(%X, %Y) 2.59/1.27 || INT(0, s(%X)) -> INT(s(0), s(%X)) 2.59/1.27 || 2.59/1.27 || R is empty. 2.59/1.27 || The set Q consists of the following terms: 2.59/1.27 || 2.59/1.27 || intlist(cons(x0, x1)) 2.59/1.27 || int(0, s(x0)) 2.59/1.27 || int(s(x0), s(x1)) 2.59/1.27 || 2.59/1.27 || We have to consider all minimal (P,Q,R)-chains. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (21) QReductionProof (EQUIVALENT) 2.59/1.27 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 2.59/1.27 || 2.59/1.27 || intlist(cons(x0, x1)) 2.59/1.27 || int(0, s(x0)) 2.59/1.27 || int(s(x0), s(x1)) 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (22) 2.59/1.27 || Obligation: 2.59/1.27 || Q DP problem: 2.59/1.27 || The TRS P consists of the following rules: 2.59/1.27 || 2.59/1.27 || INT(s(%X), s(%Y)) -> INT(%X, %Y) 2.59/1.27 || INT(0, s(%X)) -> INT(s(0), s(%X)) 2.59/1.27 || 2.59/1.27 || R is empty. 2.59/1.27 || Q is empty. 2.59/1.27 || We have to consider all minimal (P,Q,R)-chains. 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (23) QDPSizeChangeProof (EQUIVALENT) 2.59/1.27 || 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. 2.59/1.27 || 2.59/1.27 || From the DPs we obtained the following set of size-change graphs: 2.59/1.27 || *INT(s(%X), s(%Y)) -> INT(%X, %Y) 2.59/1.27 || The graph contains the following edges 1 > 1, 2 > 2 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || *INT(0, s(%X)) -> INT(s(0), s(%X)) 2.59/1.27 || The graph contains the following edges 2 >= 2 2.59/1.27 || 2.59/1.27 || 2.59/1.27 || ---------------------------------------- 2.59/1.27 || 2.59/1.27 || (24) 2.59/1.27 || YES 2.59/1.27 || 2.59/1.27 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]). 2.59/1.27 2.59/1.27 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 2.59/1.27 2.59/1.27 Dependency Pairs P_0: 2.59/1.27 2.59/1.27 0] map#(F, cons(X, Y)) =#> map#(F, Y) 2.59/1.27 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2.59/1.27 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 2.59/1.27 3] filter2#(false, F, X, Y) =#> filter#(F, Y) 2.59/1.27 2.59/1.27 Rules R_0: 2.59/1.27 2.59/1.27 intlist(nil) => nil 2.59/1.27 intlist(cons(X, Y)) => cons(s(X), intlist(Y)) 2.59/1.27 int(0, 0) => cons(0, nil) 2.59/1.27 int(0, s(X)) => cons(0, int(s(0), s(X))) 2.59/1.27 int(s(X), 0) => nil 2.59/1.27 int(s(X), s(Y)) => intlist(int(X, Y)) 2.59/1.27 map(F, nil) => nil 2.59/1.27 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 2.59/1.27 filter(F, nil) => nil 2.59/1.27 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 2.59/1.27 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 2.59/1.27 filter2(false, F, X, Y) => filter(F, Y) 2.59/1.27 2.59/1.27 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 2.59/1.27 2.59/1.27 We consider the dependency pair problem (P_0, R_0, static, formative). 2.59/1.27 2.59/1.27 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 2.59/1.27 2.59/1.27 * 0 : 0 2.59/1.27 * 1 : 2, 3 2.59/1.27 * 2 : 1 2.59/1.27 * 3 : 1 2.59/1.27 2.59/1.27 This graph has the following strongly connected components: 2.59/1.27 2.59/1.27 P_1: 2.59/1.27 2.59/1.27 map#(F, cons(X, Y)) =#> map#(F, Y) 2.59/1.27 2.59/1.27 P_2: 2.59/1.27 2.59/1.27 filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2.59/1.27 filter2#(true, F, X, Y) =#> filter#(F, Y) 2.59/1.27 filter2#(false, F, X, Y) =#> filter#(F, Y) 2.59/1.27 2.59/1.27 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). 2.59/1.27 2.59/1.27 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite. 2.59/1.27 2.59/1.27 We consider the dependency pair problem (P_2, R_0, static, formative). 2.59/1.27 2.59/1.27 We apply the subterm criterion with the following projection function: 2.59/1.27 2.59/1.27 nu(filter2#) = 4 2.59/1.27 nu(filter#) = 2 2.59/1.27 2.59/1.27 Thus, we can orient the dependency pairs as follows: 2.59/1.27 2.59/1.27 nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 2.59/1.27 nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 2.59/1.27 nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 2.59/1.27 2.59/1.27 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: 2.59/1.27 2.59/1.27 filter2#(true, F, X, Y) =#> filter#(F, Y) 2.59/1.27 filter2#(false, F, X, Y) =#> filter#(F, Y) 2.59/1.27 2.59/1.27 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite. 2.59/1.27 2.59/1.27 We consider the dependency pair problem (P_3, R_0, static, formative). 2.59/1.27 2.59/1.27 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 2.59/1.27 2.59/1.27 * 0 : 2.59/1.27 * 1 : 2.59/1.27 2.59/1.27 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 2.59/1.27 2.59/1.27 Thus, the original system is terminating if (P_1, R_0, static, formative) is finite. 2.59/1.27 2.59/1.27 We consider the dependency pair problem (P_1, R_0, static, formative). 2.59/1.27 2.59/1.27 We apply the subterm criterion with the following projection function: 2.59/1.27 2.59/1.27 nu(map#) = 2 2.59/1.27 2.59/1.27 Thus, we can orient the dependency pairs as follows: 2.59/1.27 2.59/1.27 nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 2.59/1.27 2.59/1.27 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. 2.59/1.27 2.59/1.27 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 2.59/1.27 2.59/1.27 2.59/1.27 +++ Citations +++ 2.59/1.27 2.59/1.27 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 2.59/1.27 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 2.59/1.27 [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. 2.59/1.27 EOF