2.26/1.14 YES 2.26/1.15 We consider the system theBenchmark. 2.26/1.15 2.26/1.15 Alphabet: 2.26/1.15 2.26/1.15 0 : [] --> a 2.26/1.15 cons : [c * d] --> d 2.26/1.15 false : [] --> b 2.26/1.15 filter : [c -> b * d] --> d 2.26/1.15 filter2 : [b * c -> b * c * d] --> d 2.26/1.15 map : [c -> c * d] --> d 2.26/1.15 nil : [] --> d 2.26/1.15 plus : [a * a] --> a 2.26/1.15 s : [a] --> a 2.26/1.15 times : [a * a] --> a 2.26/1.15 true : [] --> b 2.26/1.15 2.26/1.15 Rules: 2.26/1.15 2.26/1.15 times(x, plus(y, s(z))) => plus(times(x, plus(y, times(s(z), 0))), times(x, s(z))) 2.26/1.15 times(x, 0) => 0 2.26/1.15 times(x, s(y)) => plus(times(x, y), x) 2.26/1.15 plus(x, 0) => x 2.26/1.15 plus(x, s(y)) => s(plus(x, y)) 2.26/1.15 map(f, nil) => nil 2.26/1.15 map(f, cons(x, y)) => cons(f x, map(f, y)) 2.26/1.15 filter(f, nil) => nil 2.26/1.15 filter(f, cons(x, y)) => filter2(f x, f, x, y) 2.26/1.15 filter2(true, f, x, y) => cons(x, filter(f, y)) 2.26/1.15 filter2(false, f, x, y) => filter(f, y) 2.26/1.15 2.26/1.15 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 2.26/1.15 2.26/1.15 We observe that the rules contain a first-order subset: 2.26/1.15 2.26/1.15 times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) 2.26/1.15 times(X, 0) => 0 2.26/1.15 times(X, s(Y)) => plus(times(X, Y), X) 2.26/1.15 plus(X, 0) => X 2.26/1.15 plus(X, s(Y)) => s(plus(X, Y)) 2.26/1.15 2.26/1.15 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. 2.26/1.15 2.26/1.15 According to the external first-order termination prover, this system is indeed Ce-terminating: 2.26/1.15 2.26/1.15 || proof of resources/system.trs 2.26/1.15 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 2.26/1.15 || 2.26/1.15 || 2.26/1.15 || Termination w.r.t. Q of the given QTRS could be proven: 2.26/1.15 || 2.26/1.15 || (0) QTRS 2.26/1.15 || (1) DependencyPairsProof [EQUIVALENT] 2.26/1.15 || (2) QDP 2.26/1.15 || (3) DependencyGraphProof [EQUIVALENT] 2.26/1.15 || (4) AND 2.26/1.15 || (5) QDP 2.26/1.15 || (6) UsableRulesProof [EQUIVALENT] 2.26/1.15 || (7) QDP 2.26/1.15 || (8) QDPSizeChangeProof [EQUIVALENT] 2.26/1.15 || (9) YES 2.26/1.15 || (10) QDP 2.26/1.15 || (11) UsableRulesProof [EQUIVALENT] 2.26/1.15 || (12) QDP 2.26/1.15 || (13) MRRProof [EQUIVALENT] 2.26/1.15 || (14) QDP 2.26/1.15 || (15) MRRProof [EQUIVALENT] 2.26/1.15 || (16) QDP 2.26/1.15 || (17) MRRProof [EQUIVALENT] 2.26/1.15 || (18) QDP 2.26/1.15 || (19) DependencyGraphProof [EQUIVALENT] 2.26/1.15 || (20) TRUE 2.26/1.15 || 2.26/1.15 || 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (0) 2.26/1.15 || Obligation: 2.26/1.15 || Q restricted rewrite system: 2.26/1.15 || The TRS R consists of the following rules: 2.26/1.15 || 2.26/1.15 || times(%X, plus(%Y, s(%Z))) -> plus(times(%X, plus(%Y, times(s(%Z), 0))), times(%X, s(%Z))) 2.26/1.15 || times(%X, 0) -> 0 2.26/1.15 || times(%X, s(%Y)) -> plus(times(%X, %Y), %X) 2.26/1.15 || plus(%X, 0) -> %X 2.26/1.15 || plus(%X, s(%Y)) -> s(plus(%X, %Y)) 2.26/1.15 || ~PAIR(%X, %Y) -> %X 2.26/1.15 || ~PAIR(%X, %Y) -> %Y 2.26/1.15 || 2.26/1.15 || Q is empty. 2.26/1.15 || 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (1) DependencyPairsProof (EQUIVALENT) 2.26/1.15 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (2) 2.26/1.15 || Obligation: 2.26/1.15 || Q DP problem: 2.26/1.15 || The TRS P consists of the following rules: 2.26/1.15 || 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> PLUS(times(%X, plus(%Y, times(s(%Z), 0))), times(%X, s(%Z))) 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(%X, plus(%Y, times(s(%Z), 0))) 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> PLUS(%Y, times(s(%Z), 0)) 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(s(%Z), 0) 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(%X, s(%Z)) 2.26/1.15 || TIMES(%X, s(%Y)) -> PLUS(times(%X, %Y), %X) 2.26/1.15 || TIMES(%X, s(%Y)) -> TIMES(%X, %Y) 2.26/1.15 || PLUS(%X, s(%Y)) -> PLUS(%X, %Y) 2.26/1.15 || 2.26/1.15 || The TRS R consists of the following rules: 2.26/1.15 || 2.26/1.15 || times(%X, plus(%Y, s(%Z))) -> plus(times(%X, plus(%Y, times(s(%Z), 0))), times(%X, s(%Z))) 2.26/1.15 || times(%X, 0) -> 0 2.26/1.15 || times(%X, s(%Y)) -> plus(times(%X, %Y), %X) 2.26/1.15 || plus(%X, 0) -> %X 2.26/1.15 || plus(%X, s(%Y)) -> s(plus(%X, %Y)) 2.26/1.15 || ~PAIR(%X, %Y) -> %X 2.26/1.15 || ~PAIR(%X, %Y) -> %Y 2.26/1.15 || 2.26/1.15 || Q is empty. 2.26/1.15 || We have to consider all minimal (P,Q,R)-chains. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (3) DependencyGraphProof (EQUIVALENT) 2.26/1.15 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (4) 2.26/1.15 || Complex Obligation (AND) 2.26/1.15 || 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (5) 2.26/1.15 || Obligation: 2.26/1.15 || Q DP problem: 2.26/1.15 || The TRS P consists of the following rules: 2.26/1.15 || 2.26/1.15 || PLUS(%X, s(%Y)) -> PLUS(%X, %Y) 2.26/1.15 || 2.26/1.15 || The TRS R consists of the following rules: 2.26/1.15 || 2.26/1.15 || times(%X, plus(%Y, s(%Z))) -> plus(times(%X, plus(%Y, times(s(%Z), 0))), times(%X, s(%Z))) 2.26/1.15 || times(%X, 0) -> 0 2.26/1.15 || times(%X, s(%Y)) -> plus(times(%X, %Y), %X) 2.26/1.15 || plus(%X, 0) -> %X 2.26/1.15 || plus(%X, s(%Y)) -> s(plus(%X, %Y)) 2.26/1.15 || ~PAIR(%X, %Y) -> %X 2.26/1.15 || ~PAIR(%X, %Y) -> %Y 2.26/1.15 || 2.26/1.15 || Q is empty. 2.26/1.15 || We have to consider all minimal (P,Q,R)-chains. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (6) UsableRulesProof (EQUIVALENT) 2.26/1.15 || 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. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (7) 2.26/1.15 || Obligation: 2.26/1.15 || Q DP problem: 2.26/1.15 || The TRS P consists of the following rules: 2.26/1.15 || 2.26/1.15 || PLUS(%X, s(%Y)) -> PLUS(%X, %Y) 2.26/1.15 || 2.26/1.15 || R is empty. 2.26/1.15 || Q is empty. 2.26/1.15 || We have to consider all minimal (P,Q,R)-chains. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (8) QDPSizeChangeProof (EQUIVALENT) 2.26/1.15 || 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. 2.26/1.15 || 2.26/1.15 || From the DPs we obtained the following set of size-change graphs: 2.26/1.15 || *PLUS(%X, s(%Y)) -> PLUS(%X, %Y) 2.26/1.15 || The graph contains the following edges 1 >= 1, 2 > 2 2.26/1.15 || 2.26/1.15 || 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (9) 2.26/1.15 || YES 2.26/1.15 || 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (10) 2.26/1.15 || Obligation: 2.26/1.15 || Q DP problem: 2.26/1.15 || The TRS P consists of the following rules: 2.26/1.15 || 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(%X, s(%Z)) 2.26/1.15 || TIMES(%X, s(%Y)) -> TIMES(%X, %Y) 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(%X, plus(%Y, times(s(%Z), 0))) 2.26/1.15 || 2.26/1.15 || The TRS R consists of the following rules: 2.26/1.15 || 2.26/1.15 || times(%X, plus(%Y, s(%Z))) -> plus(times(%X, plus(%Y, times(s(%Z), 0))), times(%X, s(%Z))) 2.26/1.15 || times(%X, 0) -> 0 2.26/1.15 || times(%X, s(%Y)) -> plus(times(%X, %Y), %X) 2.26/1.15 || plus(%X, 0) -> %X 2.26/1.15 || plus(%X, s(%Y)) -> s(plus(%X, %Y)) 2.26/1.15 || ~PAIR(%X, %Y) -> %X 2.26/1.15 || ~PAIR(%X, %Y) -> %Y 2.26/1.15 || 2.26/1.15 || Q is empty. 2.26/1.15 || We have to consider all minimal (P,Q,R)-chains. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (11) UsableRulesProof (EQUIVALENT) 2.26/1.15 || 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. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (12) 2.26/1.15 || Obligation: 2.26/1.15 || Q DP problem: 2.26/1.15 || The TRS P consists of the following rules: 2.26/1.15 || 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(%X, s(%Z)) 2.26/1.15 || TIMES(%X, s(%Y)) -> TIMES(%X, %Y) 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(%X, plus(%Y, times(s(%Z), 0))) 2.26/1.15 || 2.26/1.15 || The TRS R consists of the following rules: 2.26/1.15 || 2.26/1.15 || times(%X, 0) -> 0 2.26/1.15 || plus(%X, 0) -> %X 2.26/1.15 || plus(%X, s(%Y)) -> s(plus(%X, %Y)) 2.26/1.15 || 2.26/1.15 || Q is empty. 2.26/1.15 || We have to consider all minimal (P,Q,R)-chains. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (13) MRRProof (EQUIVALENT) 2.26/1.15 || 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. 2.26/1.15 || 2.26/1.15 || Strictly oriented dependency pairs: 2.26/1.15 || 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(%X, s(%Z)) 2.26/1.15 || 2.26/1.15 || Strictly oriented rules of the TRS R: 2.26/1.15 || 2.26/1.15 || plus(%X, 0) -> %X 2.26/1.15 || 2.26/1.15 || Used ordering: Polynomial interpretation [POLO]: 2.26/1.15 || 2.26/1.15 || POL(0) = 0 2.26/1.15 || POL(TIMES(x_1, x_2)) = x_1 + 2*x_2 2.26/1.15 || POL(plus(x_1, x_2)) = 1 + x_1 + 2*x_2 2.26/1.15 || POL(s(x_1)) = x_1 2.26/1.15 || POL(times(x_1, x_2)) = x_1 + 2*x_2 2.26/1.15 || 2.26/1.15 || 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (14) 2.26/1.15 || Obligation: 2.26/1.15 || Q DP problem: 2.26/1.15 || The TRS P consists of the following rules: 2.26/1.15 || 2.26/1.15 || TIMES(%X, s(%Y)) -> TIMES(%X, %Y) 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(%X, plus(%Y, times(s(%Z), 0))) 2.26/1.15 || 2.26/1.15 || The TRS R consists of the following rules: 2.26/1.15 || 2.26/1.15 || times(%X, 0) -> 0 2.26/1.15 || plus(%X, s(%Y)) -> s(plus(%X, %Y)) 2.26/1.15 || 2.26/1.15 || Q is empty. 2.26/1.15 || We have to consider all minimal (P,Q,R)-chains. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (15) MRRProof (EQUIVALENT) 2.26/1.15 || 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. 2.26/1.15 || 2.26/1.15 || Strictly oriented dependency pairs: 2.26/1.15 || 2.26/1.15 || TIMES(%X, s(%Y)) -> TIMES(%X, %Y) 2.26/1.15 || 2.26/1.15 || 2.26/1.15 || Used ordering: Polynomial interpretation [POLO]: 2.26/1.15 || 2.26/1.15 || POL(0) = 0 2.26/1.15 || POL(TIMES(x_1, x_2)) = x_1 + 2*x_2 2.26/1.15 || POL(plus(x_1, x_2)) = 1 + 2*x_1 + x_2 2.26/1.15 || POL(s(x_1)) = 1 + x_1 2.26/1.15 || POL(times(x_1, x_2)) = x_1 + x_2 2.26/1.15 || 2.26/1.15 || 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (16) 2.26/1.15 || Obligation: 2.26/1.15 || Q DP problem: 2.26/1.15 || The TRS P consists of the following rules: 2.26/1.15 || 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(%X, plus(%Y, times(s(%Z), 0))) 2.26/1.15 || 2.26/1.15 || The TRS R consists of the following rules: 2.26/1.15 || 2.26/1.15 || times(%X, 0) -> 0 2.26/1.15 || plus(%X, s(%Y)) -> s(plus(%X, %Y)) 2.26/1.15 || 2.26/1.15 || Q is empty. 2.26/1.15 || We have to consider all minimal (P,Q,R)-chains. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (17) MRRProof (EQUIVALENT) 2.26/1.15 || 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. 2.26/1.15 || 2.26/1.15 || 2.26/1.15 || Strictly oriented rules of the TRS R: 2.26/1.15 || 2.26/1.15 || plus(%X, s(%Y)) -> s(plus(%X, %Y)) 2.26/1.15 || 2.26/1.15 || Used ordering: Polynomial interpretation [POLO]: 2.26/1.15 || 2.26/1.15 || POL(0) = 0 2.26/1.15 || POL(TIMES(x_1, x_2)) = x_1 + 2*x_2 2.26/1.15 || POL(plus(x_1, x_2)) = 2*x_1 + 2*x_2 2.26/1.15 || POL(s(x_1)) = 2 + x_1 2.26/1.15 || POL(times(x_1, x_2)) = x_1 + x_2 2.26/1.15 || 2.26/1.15 || 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (18) 2.26/1.15 || Obligation: 2.26/1.15 || Q DP problem: 2.26/1.15 || The TRS P consists of the following rules: 2.26/1.15 || 2.26/1.15 || TIMES(%X, plus(%Y, s(%Z))) -> TIMES(%X, plus(%Y, times(s(%Z), 0))) 2.26/1.15 || 2.26/1.15 || The TRS R consists of the following rules: 2.26/1.15 || 2.26/1.15 || times(%X, 0) -> 0 2.26/1.15 || 2.26/1.15 || Q is empty. 2.26/1.15 || We have to consider all minimal (P,Q,R)-chains. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (19) DependencyGraphProof (EQUIVALENT) 2.26/1.15 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 2.26/1.15 || ---------------------------------------- 2.26/1.15 || 2.26/1.15 || (20) 2.26/1.15 || TRUE 2.26/1.15 || 2.26/1.15 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]). 2.26/1.15 2.26/1.15 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 2.26/1.15 2.26/1.15 Dependency Pairs P_0: 2.26/1.15 2.26/1.15 0] map#(F, cons(X, Y)) =#> map#(F, Y) 2.26/1.15 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2.26/1.15 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 2.26/1.15 3] filter2#(false, F, X, Y) =#> filter#(F, Y) 2.26/1.15 2.26/1.15 Rules R_0: 2.26/1.15 2.26/1.15 times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) 2.26/1.15 times(X, 0) => 0 2.26/1.15 times(X, s(Y)) => plus(times(X, Y), X) 2.26/1.15 plus(X, 0) => X 2.26/1.15 plus(X, s(Y)) => s(plus(X, Y)) 2.26/1.15 map(F, nil) => nil 2.26/1.15 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 2.26/1.15 filter(F, nil) => nil 2.26/1.15 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 2.26/1.15 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 2.26/1.15 filter2(false, F, X, Y) => filter(F, Y) 2.26/1.15 2.26/1.15 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 2.26/1.15 2.26/1.15 We consider the dependency pair problem (P_0, R_0, static, formative). 2.26/1.15 2.26/1.15 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 2.26/1.15 2.26/1.15 * 0 : 0 2.26/1.15 * 1 : 2, 3 2.26/1.15 * 2 : 1 2.26/1.15 * 3 : 1 2.26/1.15 2.26/1.15 This graph has the following strongly connected components: 2.26/1.15 2.26/1.15 P_1: 2.26/1.15 2.26/1.15 map#(F, cons(X, Y)) =#> map#(F, Y) 2.26/1.15 2.26/1.15 P_2: 2.26/1.15 2.26/1.15 filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2.26/1.15 filter2#(true, F, X, Y) =#> filter#(F, Y) 2.26/1.15 filter2#(false, F, X, Y) =#> filter#(F, Y) 2.26/1.15 2.26/1.15 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). 2.26/1.15 2.26/1.15 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite. 2.26/1.15 2.26/1.15 We consider the dependency pair problem (P_2, R_0, static, formative). 2.26/1.15 2.26/1.15 We apply the subterm criterion with the following projection function: 2.26/1.15 2.26/1.15 nu(filter2#) = 4 2.26/1.15 nu(filter#) = 2 2.26/1.15 2.26/1.15 Thus, we can orient the dependency pairs as follows: 2.26/1.15 2.26/1.15 nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 2.26/1.15 nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 2.26/1.15 nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 2.26/1.15 2.26/1.15 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: 2.26/1.15 2.26/1.15 filter2#(true, F, X, Y) =#> filter#(F, Y) 2.26/1.15 filter2#(false, F, X, Y) =#> filter#(F, Y) 2.26/1.15 2.26/1.15 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite. 2.26/1.15 2.26/1.15 We consider the dependency pair problem (P_3, R_0, static, formative). 2.26/1.15 2.26/1.15 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 2.26/1.15 2.26/1.15 * 0 : 2.26/1.15 * 1 : 2.26/1.15 2.26/1.15 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 2.26/1.15 2.26/1.15 Thus, the original system is terminating if (P_1, R_0, static, formative) is finite. 2.26/1.15 2.26/1.15 We consider the dependency pair problem (P_1, R_0, static, formative). 2.26/1.15 2.26/1.15 We apply the subterm criterion with the following projection function: 2.26/1.15 2.26/1.15 nu(map#) = 2 2.26/1.15 2.26/1.15 Thus, we can orient the dependency pairs as follows: 2.26/1.15 2.26/1.15 nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 2.26/1.15 2.26/1.15 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. 2.26/1.15 2.26/1.15 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 2.26/1.15 2.26/1.15 2.26/1.15 +++ Citations +++ 2.26/1.15 2.26/1.15 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 2.26/1.15 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 2.26/1.15 [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. 2.26/1.15 EOF