We consider the system Applicative_AG01_innermost__#4.26. Alphabet: 0 : [] --> b cons : [c * d] --> d false : [] --> a filter : [c -> a * d] --> d filter2 : [a * c -> a * c * d] --> d if : [a * b * b] --> b le : [b * b] --> a map : [c -> c * d] --> d minus : [b * b] --> b nil : [] --> d p : [b] --> b s : [b] --> b true : [] --> a Rules: p(0) => 0 p(s(x)) => x le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) minus(x, y) => if(le(x, y), x, y) if(true, x, y) => 0 if(false, x, y) => s(minus(p(x), 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: p(0) => 0 p(s(X)) => X le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(X, Y) => if(le(X, Y), X, Y) if(true, X, Y) => 0 if(false, X, Y) => s(minus(p(X), 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 mutermprover, this system is indeed terminating: || || Problem 1: || || (VAR %X %Y) || (RULES || if(false,%X,%Y) -> s(minus(p(%X),%Y)) || if(true,%X,%Y) -> 0 || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(%X,%Y) -> if(le(%X,%Y),%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || ) || || Problem 1: || || Innermost Equivalent Processor: || -> Rules: || if(false,%X,%Y) -> s(minus(p(%X),%Y)) || if(true,%X,%Y) -> 0 || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(%X,%Y) -> if(le(%X,%Y),%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. || || || Problem 1: || || Dependency Pairs Processor: || -> Pairs: || IF(false,%X,%Y) -> MINUS(p(%X),%Y) || IF(false,%X,%Y) -> P(%X) || LE(s(%X),s(%Y)) -> LE(%X,%Y) || MINUS(%X,%Y) -> IF(le(%X,%Y),%X,%Y) || MINUS(%X,%Y) -> LE(%X,%Y) || -> Rules: || if(false,%X,%Y) -> s(minus(p(%X),%Y)) || if(true,%X,%Y) -> 0 || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(%X,%Y) -> if(le(%X,%Y),%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || || Problem 1: || || SCC Processor: || -> Pairs: || IF(false,%X,%Y) -> MINUS(p(%X),%Y) || IF(false,%X,%Y) -> P(%X) || LE(s(%X),s(%Y)) -> LE(%X,%Y) || MINUS(%X,%Y) -> IF(le(%X,%Y),%X,%Y) || MINUS(%X,%Y) -> LE(%X,%Y) || -> Rules: || if(false,%X,%Y) -> s(minus(p(%X),%Y)) || if(true,%X,%Y) -> 0 || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(%X,%Y) -> if(le(%X,%Y),%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || ->Strongly Connected Components: || ->->Cycle: || ->->-> Pairs: || LE(s(%X),s(%Y)) -> LE(%X,%Y) || ->->-> Rules: || if(false,%X,%Y) -> s(minus(p(%X),%Y)) || if(true,%X,%Y) -> 0 || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(%X,%Y) -> if(le(%X,%Y),%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || ->->Cycle: || ->->-> Pairs: || IF(false,%X,%Y) -> MINUS(p(%X),%Y) || MINUS(%X,%Y) -> IF(le(%X,%Y),%X,%Y) || ->->-> Rules: || if(false,%X,%Y) -> s(minus(p(%X),%Y)) || if(true,%X,%Y) -> 0 || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(%X,%Y) -> if(le(%X,%Y),%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || || || The problem is decomposed in 2 subproblems. || || Problem 1.1: || || Subterm Processor: || -> Pairs: || LE(s(%X),s(%Y)) -> LE(%X,%Y) || -> Rules: || if(false,%X,%Y) -> s(minus(p(%X),%Y)) || if(true,%X,%Y) -> 0 || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(%X,%Y) -> if(le(%X,%Y),%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || ->Projection: || pi(LE) = 1 || || Problem 1.1: || || SCC Processor: || -> Pairs: || Empty || -> Rules: || if(false,%X,%Y) -> s(minus(p(%X),%Y)) || if(true,%X,%Y) -> 0 || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(%X,%Y) -> if(le(%X,%Y),%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || ->Strongly Connected Components: || There is no strongly connected component || || The problem is finite. || || Problem 1.2: || || Reduction Pairs Processor: || -> Pairs: || IF(false,%X,%Y) -> MINUS(p(%X),%Y) || MINUS(%X,%Y) -> IF(le(%X,%Y),%X,%Y) || -> Rules: || if(false,%X,%Y) -> s(minus(p(%X),%Y)) || if(true,%X,%Y) -> 0 || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(%X,%Y) -> if(le(%X,%Y),%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || -> Usable rules: || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || ->Interpretation type: || Linear || ->Coefficients: || All rationals || ->Dimension: || 1 || ->Bound: || 2 || ->Interpretation: || || [le](X1,X2) = 1/2.X1 || [p](X) = 1/2.X || [0] = 0 || [false] = 1/2 || [s](X) = 2.X + 2 || [true] = 0 || [IF](X1,X2,X3) = X1 + 1/2.X2 + 1/2.X3 + 1/2 || [MINUS](X1,X2) = X1 + 1/2.X2 + 1/2 || || Problem 1.2: || || SCC Processor: || -> Pairs: || MINUS(%X,%Y) -> IF(le(%X,%Y),%X,%Y) || -> Rules: || if(false,%X,%Y) -> s(minus(p(%X),%Y)) || if(true,%X,%Y) -> 0 || le(0,%X) -> true || le(s(%X),0) -> false || le(s(%X),s(%Y)) -> le(%X,%Y) || minus(%X,%Y) -> if(le(%X,%Y),%X,%Y) || p(0) -> 0 || p(s(%X)) -> %X || ->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] map#(F, cons(X, Y)) =#> map#(F, Y) 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 3] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: p(0) => 0 p(s(X)) => X le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(X, Y) => if(le(X, Y), X, Y) if(true, X, Y) => 0 if(false, X, Y) => s(minus(p(X), 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, 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 : 2, 3 * 2 : 1 * 3 : 1 This graph has the following strongly connected components: P_1: map#(F, cons(X, Y)) =#> map#(F, Y) P_2: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) 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). Thus, the original system is terminating if each of (P_1, R_0, computable, formative) and (P_2, R_0, computable, formative) is finite. We consider the dependency pair problem (P_2, R_0, computable, formative). We apply the subterm criterion with the following projection function: nu(filter2#) = 4 nu(filter#) = 2 Thus, we can orient the dependency pairs as follows: nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, computable, f) by (P_3, R_0, computable, f), where P_3 contains: filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if each of (P_1, R_0, computable, formative) and (P_3, R_0, computable, formative) is finite. We consider the dependency pair problem (P_3, 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 : * 1 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 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(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_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.