We consider the system Applicative_first_order_05__hydra. Alphabet: 0 : [] --> a cons : [c * c] --> c copy : [a * c * c] --> c f : [c] --> c false : [] --> b filter : [c -> b * c] --> c filter2 : [b * c -> b * c * c] --> c map : [c -> c * c] --> c n : [] --> a nil : [] --> c s : [a] --> a true : [] --> b Rules: f(cons(nil, x)) => x f(cons(f(cons(nil, x)), y)) => copy(n, x, y) copy(0, x, y) => f(y) copy(s(x), y, z) => copy(x, y, cons(f(y), z)) map(g, nil) => nil map(g, cons(x, y)) => cons(g x, map(g, y)) filter(g, nil) => nil filter(g, cons(x, y)) => filter2(g x, g, x, y) filter2(true, g, x, y) => cons(x, filter(g, y)) filter2(false, g, x, y) => filter(g, 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: f(cons(nil, X)) => X f(cons(f(cons(nil, X)), Y)) => copy(n, X, Y) copy(0, X, Y) => f(Y) copy(s(X), Y, Z) => copy(X, Y, cons(f(Y), Z)) 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) QDP || (5) QDPSizeChangeProof [EQUIVALENT] || (6) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || f(cons(nil, %X)) -> %X || f(cons(f(cons(nil, %X)), %Y)) -> copy(n, %X, %Y) || copy(0, %X, %Y) -> f(%Y) || copy(s(%X), %Y, %Z) -> copy(%X, %Y, cons(f(%Y), %Z)) || ~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: || || F(cons(f(cons(nil, %X)), %Y)) -> COPY(n, %X, %Y) || COPY(0, %X, %Y) -> F(%Y) || COPY(s(%X), %Y, %Z) -> COPY(%X, %Y, cons(f(%Y), %Z)) || COPY(s(%X), %Y, %Z) -> F(%Y) || || The TRS R consists of the following rules: || || f(cons(nil, %X)) -> %X || f(cons(f(cons(nil, %X)), %Y)) -> copy(n, %X, %Y) || copy(0, %X, %Y) -> f(%Y) || copy(s(%X), %Y, %Z) -> copy(%X, %Y, cons(f(%Y), %Z)) || ~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 1 SCC with 3 less nodes. || ---------------------------------------- || || (4) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || COPY(s(%X), %Y, %Z) -> COPY(%X, %Y, cons(f(%Y), %Z)) || || The TRS R consists of the following rules: || || f(cons(nil, %X)) -> %X || f(cons(f(cons(nil, %X)), %Y)) -> copy(n, %X, %Y) || copy(0, %X, %Y) -> f(%Y) || copy(s(%X), %Y, %Z) -> copy(%X, %Y, cons(f(%Y), %Z)) || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (5) 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: || *COPY(s(%X), %Y, %Z) -> COPY(%X, %Y, cons(f(%Y), %Z)) || The graph contains the following edges 1 > 1, 2 >= 2 || || || ---------------------------------------- || || (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: f(cons(nil, X)) => X f(cons(f(cons(nil, X)), Y)) => copy(n, X, Y) copy(0, X, Y) => f(Y) copy(s(X), Y, Z) => copy(X, Y, cons(f(Y), Z)) 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 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) f(cons(nil, X)) >= X f(cons(f(cons(nil, X)), Y)) >= copy(n, X, Y) copy(0, X, Y) >= f(Y) copy(s(X), Y, Z) >= copy(X, Y, cons(f(Y), Z)) 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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 cons = \y0y1.3 + y1 + 2y0 copy = \y0y1y2.3 + y2 + 3y0 + y0y1 f = \y0.y0 false = 3 filter = \G0y1.2y1 + G0(y1) + y1G0(y1) filter2 = \y0G1y2y3.3 + 2y2 + 2y3 + G1(y3) + y3G1(y3) filter2# = \y0G1y2y3.y3G1(y3) filter# = \G0y1.y1G0(y1) map = \G0y1.2y1 + 2G0(0) + 3y1G0(y1) map# = \G0y1.3 + y1 + 2G0(y1) n = 0 nil = 0 s = \y0.3 + y0 true = 3 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 6 + x2 + 2x1 + 2F0(3 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 6 + x2 + 2x1 + 2F0(3 + x2 + 2x1) > 3 + x2 + 2F0(x2) = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2x1F0(3 + x2 + 2x1) + 3F0(3 + x2 + 2x1) + x2F0(3 + x2 + 2x1) >= x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2x1F0(3 + x2 + 2x1) + 3F0(3 + x2 + 2x1) + x2F0(3 + x2 + 2x1) >= F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = x2F0(x2) >= x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = x2F0(x2) >= x2F0(x2) = [[filter#(_F0, _x2)]] [[f(cons(nil, _x0))]] = 3 + x0 >= x0 = [[_x0]] [[f(cons(f(cons(nil, _x0)), _x1))]] = 9 + x1 + 2x0 >= 3 + x1 = [[copy(n, _x0, _x1)]] [[copy(0, _x0, _x1)]] = 12 + x1 + 3x0 >= x1 = [[f(_x1)]] [[copy(s(_x0), _x1, _x2)]] = 12 + x2 + 3x0 + 3x1 + x0x1 >= 6 + x2 + 2x1 + 3x0 + x0x1 = [[copy(_x0, _x1, cons(f(_x1), _x2))]] [[map(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 4x1 + 2F0(0) + 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) >= 3 + 2x2 + 2F0(0) + 3x2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = F0(0) >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 4x1 + 2x1F0(3 + x2 + 2x1) + 4F0(3 + x2 + 2x1) + x2F0(3 + x2 + 2x1) >= 3 + 2x1 + 2x2 + F0(x2) + x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 3 + 2x1 + 2x2 + F0(x2) + x2F0(x2) >= 3 + 2x1 + 2x2 + F0(x2) + x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 3 + 2x1 + 2x2 + F0(x2) + x2F0(x2) >= 2x2 + F0(x2) + x2F0(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_0, minimal, formative) by (P_1, R_0, minimal, formative), where P_1 consists of: 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) 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 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: 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) f(cons(nil, X)) >= X f(cons(f(cons(nil, X)), Y)) >= copy(n, X, Y) copy(0, X, Y) >= f(Y) copy(s(X), Y, Z) >= copy(X, Y, cons(f(Y), Z)) 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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 cons = \y0y1.1 + y0 + y1 copy = \y0y1y2.y2 + 2y0 + y0y1 f = \y0.y0 false = 3 filter = \G0y1.y1 + 2G0(y1) + y1G0(y1) filter2 = \y0G1y2y3.1 + y2 + y3 + 2G1(y3) + y3G1(y3) filter2# = \y0G1y2y3.y3 + G1(y3) filter# = \G0y1.y1 + G0(y1) map = \G0y1.y1 + y1G0(y1) n = 0 nil = 0 s = \y0.3 + 2y0 true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + F0(1 + x1 + x2) > x2 + F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + F0(1 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = x2 + F0(x2) >= x2 + F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = x2 + F0(x2) >= x2 + F0(x2) = [[filter#(_F0, _x2)]] [[f(cons(nil, _x0))]] = 1 + x0 >= x0 = [[_x0]] [[f(cons(f(cons(nil, _x0)), _x1))]] = 2 + x0 + x1 >= x1 = [[copy(n, _x0, _x1)]] [[copy(0, _x0, _x1)]] = 6 + x1 + 3x0 >= x1 = [[f(_x1)]] [[copy(s(_x0), _x1, _x2)]] = 6 + x2 + 2x0x1 + 3x1 + 4x0 >= 1 + x1 + x2 + 2x0 + x0x1 = [[copy(_x0, _x1, cons(f(_x1), _x2))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) >= 1 + x2 + x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + 3F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) >= 1 + x1 + x2 + 2F0(x2) + x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 1 + x1 + x2 + 2F0(x2) + x2F0(x2) >= 1 + x1 + x2 + 2F0(x2) + x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 1 + x1 + x2 + 2F0(x2) + x2F0(x2) >= x2 + 2F0(x2) + x2F0(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_1, R_0, minimal, formative) by (P_2, R_0, minimal, formative), where P_2 consists of: filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, 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. 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.