We consider the system Applicative_first_order_05__#3.38. Alphabet: 0 : [] --> c cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d nil : [] --> d rev : [d] --> d rev1 : [c * d] --> c rev2 : [c * d] --> d s : [a] --> c true : [] --> b Rules: rev(nil) => nil rev(cons(x, y)) => cons(rev1(x, y), rev2(x, y)) rev1(0, nil) => 0 rev1(s(x), nil) => s(x) rev1(x, cons(y, z)) => rev1(y, z) rev2(x, nil) => nil rev2(x, cons(y, z)) => rev(cons(x, rev2(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) 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: rev(nil) => nil rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) rev1(0, nil) => 0 rev1(s(X), nil) => s(X) rev1(X, cons(Y, Z)) => rev1(Y, Z) rev2(X, nil) => nil rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) 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 nattprover, this system is indeed terminating: || Input TRS: || 1: rev(nil()) -> nil() || 2: rev(cons(PeRCenTX,PeRCenTY)) -> cons(rev1(PeRCenTX,PeRCenTY),rev2(PeRCenTX,PeRCenTY)) || 3: rev1(0(),nil()) -> 0() || 4: rev1(s(PeRCenTX),nil()) -> s(PeRCenTX) || 5: rev1(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> rev1(PeRCenTY,PeRCenTZ) || 6: rev2(PeRCenTX,nil()) -> nil() || 7: rev2(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> rev(cons(PeRCenTX,rev2(PeRCenTY,PeRCenTZ))) || Number of strict rules: 7 || Direct POLO(bPol) ... failed. || Uncurrying rev || 1: rev^1_nil() -> nil() || 2: rev^1_cons(PeRCenTX,PeRCenTY) -> cons(rev1(PeRCenTX,PeRCenTY),rev2(PeRCenTX,PeRCenTY)) || 3: rev1(0(),nil()) -> 0() || 4: rev1(s(PeRCenTX),nil()) -> s(PeRCenTX) || 5: rev1(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> rev1(PeRCenTY,PeRCenTZ) || 6: rev2(PeRCenTX,nil()) -> nil() || 7: rev2(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> rev^1_cons(PeRCenTX,rev2(PeRCenTY,PeRCenTZ)) || 8: rev(cons(_1,_2)) ->= rev^1_cons(_1,_2) || 9: rev(nil()) ->= rev^1_nil() || Number of strict rules: 7 || Direct POLO(bPol) ... failed. || Dependency Pairs: || #1: #rev^1_cons(PeRCenTX,PeRCenTY) -> #rev1(PeRCenTX,PeRCenTY) || #2: #rev^1_cons(PeRCenTX,PeRCenTY) -> #rev2(PeRCenTX,PeRCenTY) || #3: #rev(nil()) ->? #rev^1_nil() || #4: #rev2(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> #rev^1_cons(PeRCenTX,rev2(PeRCenTY,PeRCenTZ)) || #5: #rev2(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> #rev2(PeRCenTY,PeRCenTZ) || #6: #rev1(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> #rev1(PeRCenTY,PeRCenTZ) || #7: #rev(cons(_1,_2)) ->? #rev^1_cons(_1,_2) || Number of SCCs: 2, DPs: 4 || SCC { #6 } || POLO(Sum)... succeeded. || #rev w: 0 || s w: 0 || rev1 w: 0 || #rev1 w: x2 || #rev2 w: 0 || 0 w: 0 || #rev^1_nil w: 0 || nil w: 0 || rev w: 0 || cons w: x1 + x2 + 1 || #rev^1_cons w: 0 || rev^1_cons w: 0 || rev2 w: 0 || rev^1_nil w: 0 || USABLE RULES: { } || Removed DPs: #6 || Number of SCCs: 1, DPs: 3 || SCC { #2 #4 #5 } || POLO(Sum)... succeeded. || #rev w: 0 || s w: 1 || rev1 w: x1 + x2 + 1 || #rev1 w: 0 || #rev2 w: x2 || 0 w: 1 || #rev^1_nil w: 0 || nil w: 1 || rev w: 0 || cons w: x2 + 3 || #rev^1_cons w: x2 + 1 || rev^1_cons w: x2 + 3 || rev2 w: x2 || rev^1_nil w: 0 || USABLE RULES: { 2 6 7 } || Removed DPs: #2 #4 #5 || Number of SCCs: 0, DPs: 0 || 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: rev(nil) => nil rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) rev1(0, nil) => 0 rev1(s(X), nil) => s(X) rev1(X, cons(Y, Z)) => rev1(Y, Z) rev2(X, nil) => nil rev2(X, cons(Y, Z)) => rev(cons(X, rev2(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, 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.