We consider the system Applicative_first_order_05__11. Alphabet: !facminus : [a * a] --> a !facplus : [a * a] --> a !factimes : [a * a] --> a 0 : [] --> a 1 : [] --> a 2 : [] --> a D : [a] --> a cons : [c * d] --> d constant : [] --> a div : [a * a] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d ln : [a] --> a map : [c -> c * d] --> d minus : [a] --> a nil : [] --> d pow : [a * a] --> a t : [] --> a true : [] --> b Rules: D(t) => 1 D(constant) => 0 D(!facplus(x, y)) => !facplus(D(x), D(y)) D(!factimes(x, y)) => !facplus(!factimes(y, D(x)), !factimes(x, D(y))) D(!facminus(x, y)) => !facminus(D(x), D(y)) D(minus(x)) => minus(D(x)) D(div(x, y)) => !facminus(div(D(x), y), div(!factimes(x, D(y)), pow(y, 2))) D(ln(x)) => div(D(x), x) D(pow(x, y)) => !facplus(!factimes(!factimes(y, pow(x, !facminus(y, 1))), D(x)), !factimes(!factimes(pow(x, y), ln(x)), D(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: D(t) => 1 D(constant) => 0 D(!facplus(X, Y)) => !facplus(D(X), D(Y)) D(!factimes(X, Y)) => !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) D(!facminus(X, Y)) => !facminus(D(X), D(Y)) D(minus(X)) => minus(D(X)) D(div(X, Y)) => !facminus(div(D(X), Y), div(!factimes(X, D(Y)), pow(Y, 2))) D(ln(X)) => div(D(X), X) D(pow(X, Y)) => !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, 1))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(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) QTRSRRRProof [EQUIVALENT] || (2) QTRS || (3) QTRSRRRProof [EQUIVALENT] || (4) QTRS || (5) RisEmptyProof [EQUIVALENT] || (6) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || D(t) -> 1 || D(constant) -> 0 || D(!facplus(%X, %Y)) -> !facplus(D(%X), D(%Y)) || D(!factimes(%X, %Y)) -> !facplus(!factimes(%Y, D(%X)), !factimes(%X, D(%Y))) || D(!facminus(%X, %Y)) -> !facminus(D(%X), D(%Y)) || D(minus(%X)) -> minus(D(%X)) || D(div(%X, %Y)) -> !facminus(div(D(%X), %Y), div(!factimes(%X, D(%Y)), pow(%Y, 2))) || D(ln(%X)) -> div(D(%X), %X) || D(pow(%X, %Y)) -> !facplus(!factimes(!factimes(%Y, pow(%X, !facminus(%Y, 1))), D(%X)), !factimes(!factimes(pow(%X, %Y), ln(%X)), D(%Y))) || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || D/1(YES) || t/0) || 1/0) || constant/0) || 0/0) || !facplus/2(YES,YES) || !factimes/2(YES,YES) || !facminus/2(YES,YES) || minus/1)YES( || div/2(YES,YES) || pow/2(YES,YES) || 2/0) || ln/1(YES) || || Quasi precedence: || [D_1, ln_1] > 1 > [!facplus_2, !facminus_2] || [D_1, ln_1] > 0 > [!facplus_2, !facminus_2] || [D_1, ln_1] > div_2 > !factimes_2 > [!facplus_2, !facminus_2] || [D_1, ln_1] > pow_2 > !factimes_2 > [!facplus_2, !facminus_2] || [D_1, ln_1] > 2 > [!facplus_2, !facminus_2] || t > [!facplus_2, !facminus_2] || constant > 0 > [!facplus_2, !facminus_2] || || || Status: || D_1: multiset status || t: multiset status || 1: multiset status || constant: multiset status || 0: multiset status || !facplus_2: [2,1] || !factimes_2: multiset status || !facminus_2: multiset status || div_2: multiset status || pow_2: multiset status || 2: multiset status || ln_1: multiset status || || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || D(t) -> 1 || D(constant) -> 0 || D(!facplus(%X, %Y)) -> !facplus(D(%X), D(%Y)) || D(!factimes(%X, %Y)) -> !facplus(!factimes(%Y, D(%X)), !factimes(%X, D(%Y))) || D(!facminus(%X, %Y)) -> !facminus(D(%X), D(%Y)) || D(div(%X, %Y)) -> !facminus(div(D(%X), %Y), div(!factimes(%X, D(%Y)), pow(%Y, 2))) || D(ln(%X)) -> div(D(%X), %X) || D(pow(%X, %Y)) -> !facplus(!factimes(!factimes(%Y, pow(%X, !facminus(%Y, 1))), D(%X)), !factimes(!factimes(pow(%X, %Y), ln(%X)), D(%Y))) || || || || || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || D(minus(%X)) -> minus(D(%X)) || || Q is empty. || || ---------------------------------------- || || (3) QTRSRRRProof (EQUIVALENT) || Used ordering: || Knuth-Bendix order [KBO] with precedence:D_1 > minus_1 || || and weight map: || || D_1=2 || minus_1=1 || || The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || D(minus(%X)) -> minus(D(%X)) || || || || || ---------------------------------------- || || (4) || Obligation: || Q restricted rewrite system: || R is empty. || Q is empty. || || ---------------------------------------- || || (5) RisEmptyProof (EQUIVALENT) || The TRS R is empty. Hence, termination is trivially proven. || ---------------------------------------- || || (6) || YES || We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs. After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] map#(F, cons(X, Y)) =#> F(X) 1] map#(F, cons(X, Y)) =#> map#(F, Y) 2] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 3] filter#(F, cons(X, Y)) =#> F(X) 4] filter2#(true, F, X, Y) =#> filter#(F, Y) 5] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: D(t) => 1 D(constant) => 0 D(!facplus(X, Y)) => !facplus(D(X), D(Y)) D(!factimes(X, Y)) => !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) D(!facminus(X, Y)) => !facminus(D(X), D(Y)) D(minus(X)) => minus(D(X)) D(div(X, Y)) => !facminus(div(D(X), Y), div(!factimes(X, D(Y)), pow(Y, 2))) D(ln(X)) => div(D(X), X) D(pow(X, Y)) => !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, 1))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(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). The formative rules of (P_0, R_0) are R_1 ::= map(F, cons(X, Y)) => cons(F X, map(F, Y)) 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) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative). Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70]. Thus, we must orient: map#(F, cons(X, Y)) >? F(X) map#(F, cons(X, Y)) >? map#(F, Y) filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) >? F(X) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.2 + y1 + 2y0 false = 3 filter = \G0y1.y1 + 2y1G0(y1) filter2 = \y0G1y2y3.2 + y3 + 2y2 + 2y3G1(y3) filter2# = \y0G1y2y3.y3 + 2y3G1(y3) + 2G1(y3) filter# = \G0y1.y1 + 2y1G0(y1) + 2G0(y1) map = \G0y1.2y1 + y1G0(y1) map# = \G0y1.3 + 2G0(y1) + y1G0(y1) true = 3 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) >= 3 + 2F0(x2) + x2F0(x2) = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 6F0(2 + x2 + 2x1) > x2 + 2x2F0(x2) + 2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 6F0(2 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = x2 + 2x2F0(x2) + 2F0(x2) >= x2 + 2x2F0(x2) + 2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = x2 + 2x2F0(x2) + 2F0(x2) >= x2 + 2x2F0(x2) + 2F0(x2) = [[filter#(_F0, _x2)]] [[map(_F0, cons(_x1, _x2))]] = 4 + 2x2 + 4x1 + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) >= 2 + 2x2 + x2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) >= 2 + x2 + 2x1 + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + x2 + 2x1 + 2x2F0(x2) >= 2 + x2 + 2x1 + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + x2 + 2x1 + 2x2F0(x2) >= x2 + 2x2F0(x2) = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_1, minimal, formative) by (P_1, R_1, minimal, formative), where P_1 consists of: map#(F, cons(X, Y)) =#> map#(F, Y) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_1, 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 : This graph has the following strongly connected components: P_2: map#(F, cons(X, Y)) =#> map#(F, Y) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_1, R_1, m, f) by (P_2, R_1, m, f). Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_1, 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 [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_1, minimal, f) by ({}, R_1, 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 +++ [FuhKop19] C. Fuhs, and C. Kop. A static higher-order dependency pair framework. In Proceedings of ESOP 2019, 2019. [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.