We consider the system zipWith. Alphabet: 0 : [] --> nat cons : [nat * list] --> list false : [] --> bool gcd : [nat * nat] --> nat gcdlists : [list * list] --> list if : [bool * nat * nat] --> nat le : [nat * nat] --> bool minus : [nat * nat] --> nat nil : [] --> list s : [nat] --> nat true : [] --> bool zipWith : [nat -> nat -> nat * list * list] --> list Rules: le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) minus(x, 0) => x minus(s(x), s(y)) => minus(x, y) gcd(0, x) => 0 gcd(s(x), 0) => 0 gcd(s(x), s(y)) => if(le(y, x), s(x), s(y)) if(true, s(x), s(y)) => gcd(minus(x, y), s(y)) if(false, s(x), s(y)) => gcd(minus(y, x), s(x)) zipWith(f, x, nil) => nil zipWith(f, nil, x) => nil zipWith(f, cons(x, y), cons(z, u)) => cons(f x z, zipWith(f, y, u)) gcdlists(x, y) => zipWith(/\z./\u.gcd(z, u), x, 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: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) gcd(0, X) => 0 gcd(s(X), 0) => 0 gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) if(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) 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 mutermprover, this system is indeed Ce-terminating: || || Problem 1: || || (VAR %X %Y) || (RULES || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || ) || || Problem 1: || || Dependency Pairs Processor: || -> Pairs: || GCD(s(%X),s(%Y)) -> IF(le(%Y,%X),s(%X),s(%Y)) || GCD(s(%X),s(%Y)) -> LE(%Y,%X) || IF(false,s(%X),s(%Y)) -> GCD(minus(%Y,%X),s(%X)) || IF(false,s(%X),s(%Y)) -> MINUS(%Y,%X) || IF(true,s(%X),s(%Y)) -> GCD(minus(%X,%Y),s(%Y)) || IF(true,s(%X),s(%Y)) -> MINUS(%X,%Y) || LE(s(%X),s(%Y)) -> LE(%X,%Y) || MINUS(s(%X),s(%Y)) -> MINUS(%X,%Y) || -> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || || Problem 1: || || SCC Processor: || -> Pairs: || GCD(s(%X),s(%Y)) -> IF(le(%Y,%X),s(%X),s(%Y)) || GCD(s(%X),s(%Y)) -> LE(%Y,%X) || IF(false,s(%X),s(%Y)) -> GCD(minus(%Y,%X),s(%X)) || IF(false,s(%X),s(%Y)) -> MINUS(%Y,%X) || IF(true,s(%X),s(%Y)) -> GCD(minus(%X,%Y),s(%Y)) || IF(true,s(%X),s(%Y)) -> MINUS(%X,%Y) || LE(s(%X),s(%Y)) -> LE(%X,%Y) || MINUS(s(%X),s(%Y)) -> MINUS(%X,%Y) || -> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || ->Strongly Connected Components: || ->->Cycle: || ->->-> Pairs: || MINUS(s(%X),s(%Y)) -> MINUS(%X,%Y) || ->->-> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || ->->Cycle: || ->->-> Pairs: || LE(s(%X),s(%Y)) -> LE(%X,%Y) || ->->-> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || ->->Cycle: || ->->-> Pairs: || GCD(s(%X),s(%Y)) -> IF(le(%Y,%X),s(%X),s(%Y)) || IF(false,s(%X),s(%Y)) -> GCD(minus(%Y,%X),s(%X)) || IF(true,s(%X),s(%Y)) -> GCD(minus(%X,%Y),s(%Y)) || ->->-> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || || || The problem is decomposed in 3 subproblems. || || Problem 1.1: || || Subterm Processor: || -> Pairs: || MINUS(s(%X),s(%Y)) -> MINUS(%X,%Y) || -> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || ->Projection: || pi(MINUS) = 1 || || Problem 1.1: || || SCC Processor: || -> Pairs: || Empty || -> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || ->Strongly Connected Components: || There is no strongly connected component || || The problem is finite. || || Problem 1.2: || || Subterm Processor: || -> Pairs: || LE(s(%X),s(%Y)) -> LE(%X,%Y) || -> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || ->Projection: || pi(LE) = 1 || || Problem 1.2: || || SCC Processor: || -> Pairs: || Empty || -> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || ->Strongly Connected Components: || There is no strongly connected component || || The problem is finite. || || Problem 1.3: || || Reduction Pair Processor: || -> Pairs: || GCD(s(%X),s(%Y)) -> IF(le(%Y,%X),s(%X),s(%Y)) || IF(false,s(%X),s(%Y)) -> GCD(minus(%Y,%X),s(%X)) || IF(true,s(%X),s(%Y)) -> GCD(minus(%X,%Y),s(%Y)) || -> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || -> Usable rules: || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ->Interpretation type: || Linear || ->Coefficients: || Natural Numbers || ->Dimension: || 1 || ->Bound: || 2 || ->Interpretation: || || [le](X1,X2) = X1 + 2.X2 + 2 || [minus](X1,X2) = 2.X1 || [0] = 2 || [false] = 1 || [s](X) = 2.X + 2 || [true] = 2 || [GCD](X1,X2) = 2.X1 + 2.X2 + 2 || [IF](X1,X2,X3) = 2.X2 + 2.X3 + 1 || || Problem 1.3: || || SCC Processor: || -> Pairs: || IF(false,s(%X),s(%Y)) -> GCD(minus(%Y,%X),s(%X)) || IF(true,s(%X),s(%Y)) -> GCD(minus(%X,%Y),s(%Y)) || -> Rules: || gcd(0,%X) -> 0 || gcd(s(%X),0) -> 0 || gcd(s(%X),s(%Y)) -> if(le(%Y,%X),s(%X),s(%Y)) || if(false,s(%X),s(%Y)) -> gcd(minus(%Y,%X),s(%X)) || if(true,s(%X),s(%Y)) -> gcd(minus(%X,%Y),s(%Y)) || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(s(%X),s(%Y)) -> minus(%X,%Y) || minus(%X,0) -> %X || ~PAIR(%X,%Y) -> %X || ~PAIR(%X,%Y) -> %Y || ->Strongly Connected Components: || There is no strongly connected component || || The problem is finite. || 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 [FuhKop19]). We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] zipWith#(F, cons(X, Y), cons(Z, U)) =#> zipWith#(F, Y, U) 1] gcdlists#(X, Y) =#> zipWith#(/\x./\y.gcd(x, y), X, Y) 2] gcdlists#(X, Y) =#> gcd#(Z, U) Rules R_0: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) gcd(0, X) => 0 gcd(s(X), 0) => 0 gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) if(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) zipWith(F, X, nil) => nil zipWith(F, nil, X) => nil zipWith(F, cons(X, Y), cons(Z, U)) => cons(F X Z, zipWith(F, Y, U)) gcdlists(X, Y) => zipWith(/\x./\y.gcd(x, y), X, Y) Thus, the original system is terminating if (P_0, R_0, computable, formative) is finite. We consider the dependency pair problem (P_0, R_0, computable, 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 : 0 * 2 : This graph has the following strongly connected components: P_1: zipWith#(F, cons(X, Y), cons(Z, U)) =#> zipWith#(F, Y, U) 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). Thus, the original system is terminating if (P_1, R_0, computable, formative) is finite. We consider the dependency pair problem (P_1, R_0, computable, formative). We apply the subterm criterion with the following projection function: nu(zipWith#) = 2 Thus, we can orient the dependency pairs as follows: nu(zipWith#(F, cons(X, Y), cons(Z, U))) = cons(X, Y) |> Y = nu(zipWith#(F, Y, U)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, computable, f) by ({}, R_0, computable, 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. [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.