YES We show the termination of the TRS R: app(app(append(),nil()),l) -> l app(app(append(),app(app(cons(),h),t)),l) -> app(app(cons(),h),app(app(append(),t),l)) app(app(map(),f),nil()) -> nil() app(app(map(),f),app(app(cons(),h),t)) -> app(app(cons(),app(f,h)),app(app(map(),f),t)) app(app(append(),app(app(append(),l1),l2)),l3) -> app(app(append(),l1),app(app(append(),l2),l3)) app(app(map(),f),app(app(append(),l1),l2)) -> app(app(append(),app(app(map(),f),l1)),app(app(map(),f),l2)) Instead we prove termination of the uncurried system: app(append_0(),y) -> append_1(y) app(append_1(x1),y) -> append_2(x1,y) app(cons_0(),y) -> cons_1(y) app(cons_1(x1),y) -> cons_2(x1,y) app(map_0(),y) -> map_1(y) app(map_1(x1),y) -> map_2(x1,y) append_2(nil_0(),l) -> l append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l)) map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3)) map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(append_1(x1),y) -> append_2#(x1,y) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: append_2#(cons_2(h,t),l) -> append_2#(t,l) p4: map_2#(f,cons_2(h,t)) -> app#(f,h) p5: map_2#(f,cons_2(h,t)) -> map_2#(f,t) p6: append_2#(append_2(l1,l2),l3) -> append_2#(l1,append_2(l2,l3)) p7: append_2#(append_2(l1,l2),l3) -> append_2#(l2,l3) p8: map_2#(f,append_2(l1,l2)) -> append_2#(map_2(f,l1),map_2(f,l2)) p9: map_2#(f,append_2(l1,l2)) -> map_2#(f,l1) p10: map_2#(f,append_2(l1,l2)) -> map_2#(f,l2) and R consists of: r1: app(append_0(),y) -> append_1(y) r2: app(append_1(x1),y) -> append_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: append_2(nil_0(),l) -> l r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l)) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3)) r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2)) The estimated dependency graph contains the following SCCs: {p2, p4, p5, p9, p10} {p3, p6, p7} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(h,t)) -> app#(f,h) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: map_2#(f,append_2(l1,l2)) -> map_2#(f,l2) p4: map_2#(f,append_2(l1,l2)) -> map_2#(f,l1) p5: map_2#(f,cons_2(h,t)) -> map_2#(f,t) and R consists of: r1: app(append_0(),y) -> append_1(y) r2: app(append_1(x1),y) -> append_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: append_2(nil_0(),l) -> l r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l)) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3)) r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: model-based reduction triple (=A, >=A!, >A!): finite algebra: carrier: {0,1} interpretations: map_2#_A(0,0) = 0 map_2#_A(0,1) = 0 map_2#_A(1,0) = 0 map_2#_A(1,1) = 0 cons_2_A(0,0) = 0 cons_2_A(0,1) = 1 cons_2_A(1,0) = 0 cons_2_A(1,1) = 1 app#_A(0,0) = 0 app#_A(0,1) = 0 app#_A(1,0) = 0 app#_A(1,1) = 0 map_1_A(0) = 0 map_1_A(1) = 1 append_2_A(0,0) = 0 append_2_A(0,1) = 1 append_2_A(1,0) = 0 append_2_A(1,1) = 1 app_A(0,0) = 1 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 1 append_0_A = 0 append_1_A(0) = 1 append_1_A(1) = 1 cons_0_A = 0 cons_1_A(0) = 1 cons_1_A(1) = 1 map_0_A = 1 map_2_A(0,0) = 1 map_2_A(0,1) = 1 map_2_A(1,0) = 0 map_2_A(1,1) = 1 nil_0_A = 1 map_2#!_A(0,0) = 0 map_2#!_A(0,1) = 0 map_2#!_A(1,0) = 0 map_2#!_A(1,1) = 0 cons_2!_A(0,0) = 0 cons_2!_A(0,1) = 0 cons_2!_A(1,0) = 0 cons_2!_A(1,1) = 0 app#!_A(0,0) = 0 app#!_A(0,1) = 0 app#!_A(1,0) = 0 app#!_A(1,1) = 0 map_1!_A(0) = 0 map_1!_A(1) = 0 append_2!_A(0,0) = 0 append_2!_A(0,1) = 0 append_2!_A(1,0) = 0 append_2!_A(1,1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 append_0!_A = 0 append_1!_A(0) = 0 append_1!_A(1) = 0 cons_0!_A = 0 cons_1!_A(0) = 0 cons_1!_A(1) = 0 map_0!_A = 0 map_2!_A(0,0) = 1 map_2!_A(0,1) = 1 map_2!_A(1,0) = 1 map_2!_A(1,1) = 1 nil_0!_A = 1 partial status: pi(map_2#) = [2] pi(cons_2) = [1, 2] pi(app#) = [2] pi(map_1) = [1] pi(append_2) = [1, 2] pi(app) = [1, 2] pi(append_0) = [] pi(append_1) = [1] pi(cons_0) = [] pi(cons_1) = [1] pi(map_0) = [] pi(map_2) = [1, 2] pi(nil_0) = [] The following rules are removed: p1, p3, p4 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(map_1(x1),y) -> map_2#(x1,y) p2: map_2#(f,cons_2(h,t)) -> map_2#(f,t) and R consists of: r1: app(append_0(),y) -> append_1(y) r2: app(append_1(x1),y) -> append_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: append_2(nil_0(),l) -> l r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l)) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3)) r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2)) The estimated dependency graph contains the following SCCs: {p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(h,t)) -> map_2#(f,t) and R consists of: r1: app(append_0(),y) -> append_1(y) r2: app(append_1(x1),y) -> append_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: append_2(nil_0(),l) -> l r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l)) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3)) r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: model-based reduction triple (=A, >=A!, >A!): finite algebra: carrier: {0,1} interpretations: map_2#_A(0,0) = 0 map_2#_A(0,1) = 0 map_2#_A(1,0) = 0 map_2#_A(1,1) = 0 cons_2_A(0,0) = 0 cons_2_A(0,1) = 1 cons_2_A(1,0) = 0 cons_2_A(1,1) = 1 app_A(0,0) = 0 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 0 append_0_A = 1 append_1_A(0) = 0 append_1_A(1) = 0 append_2_A(0,0) = 0 append_2_A(0,1) = 1 append_2_A(1,0) = 0 append_2_A(1,1) = 1 cons_0_A = 1 cons_1_A(0) = 0 cons_1_A(1) = 0 map_0_A = 1 map_1_A(0) = 0 map_1_A(1) = 0 map_2_A(0,0) = 0 map_2_A(0,1) = 1 map_2_A(1,0) = 0 map_2_A(1,1) = 1 nil_0_A = 1 map_2#!_A(0,0) = 0 map_2#!_A(0,1) = 1 map_2#!_A(1,0) = 0 map_2#!_A(1,1) = 1 cons_2!_A(0,0) = 0 cons_2!_A(0,1) = 0 cons_2!_A(1,0) = 0 cons_2!_A(1,1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 append_0!_A = 1 append_1!_A(0) = 0 append_1!_A(1) = 0 append_2!_A(0,0) = 0 append_2!_A(0,1) = 0 append_2!_A(1,0) = 0 append_2!_A(1,1) = 0 cons_0!_A = 1 cons_1!_A(0) = 0 cons_1!_A(1) = 0 map_0!_A = 0 map_1!_A(0) = 0 map_1!_A(1) = 0 map_2!_A(0,0) = 1 map_2!_A(0,1) = 1 map_2!_A(1,0) = 1 map_2!_A(1,1) = 1 nil_0!_A = 1 partial status: pi(map_2#) = [2] pi(cons_2) = [1, 2] pi(app) = [1, 2] pi(append_0) = [] pi(append_1) = [1] pi(append_2) = [1, 2] pi(cons_0) = [] pi(cons_1) = [1] pi(map_0) = [] pi(map_1) = [1] pi(map_2) = [1, 2] pi(nil_0) = [] The following rules are removed: p1 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: append_2#(cons_2(h,t),l) -> append_2#(t,l) p2: append_2#(append_2(l1,l2),l3) -> append_2#(l2,l3) p3: append_2#(append_2(l1,l2),l3) -> append_2#(l1,append_2(l2,l3)) and R consists of: r1: app(append_0(),y) -> append_1(y) r2: app(append_1(x1),y) -> append_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: append_2(nil_0(),l) -> l r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l)) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3)) r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: model-based reduction triple (=A, >=A!, >A!): finite algebra: carrier: {0,1} interpretations: append_2#_A(0,0) = 0 append_2#_A(0,1) = 0 append_2#_A(1,0) = 0 append_2#_A(1,1) = 0 cons_2_A(0,0) = 0 cons_2_A(0,1) = 1 cons_2_A(1,0) = 0 cons_2_A(1,1) = 1 append_2_A(0,0) = 0 append_2_A(0,1) = 1 append_2_A(1,0) = 0 append_2_A(1,1) = 1 app_A(0,0) = 0 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 0 append_0_A = 1 append_1_A(0) = 0 append_1_A(1) = 0 cons_0_A = 1 cons_1_A(0) = 0 cons_1_A(1) = 0 map_0_A = 1 map_1_A(0) = 0 map_1_A(1) = 0 map_2_A(0,0) = 0 map_2_A(0,1) = 1 map_2_A(1,0) = 0 map_2_A(1,1) = 1 nil_0_A = 1 append_2#!_A(0,0) = 1 append_2#!_A(0,1) = 1 append_2#!_A(1,0) = 1 append_2#!_A(1,1) = 1 cons_2!_A(0,0) = 0 cons_2!_A(0,1) = 0 cons_2!_A(1,0) = 0 cons_2!_A(1,1) = 0 append_2!_A(0,0) = 0 append_2!_A(0,1) = 0 append_2!_A(1,0) = 0 append_2!_A(1,1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 append_0!_A = 0 append_1!_A(0) = 0 append_1!_A(1) = 0 cons_0!_A = 0 cons_1!_A(0) = 0 cons_1!_A(1) = 0 map_0!_A = 0 map_1!_A(0) = 0 map_1!_A(1) = 0 map_2!_A(0,0) = 1 map_2!_A(0,1) = 1 map_2!_A(1,0) = 1 map_2!_A(1,1) = 1 nil_0!_A = 1 partial status: pi(append_2#) = [1] pi(cons_2) = [1, 2] pi(append_2) = [1, 2] pi(app) = [1, 2] pi(append_0) = [] pi(append_1) = [1] pi(cons_0) = [] pi(cons_1) = [1] pi(map_0) = [] pi(map_1) = [1] pi(map_2) = [1, 2] pi(nil_0) = [] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: append_2#(cons_2(h,t),l) -> append_2#(t,l) p2: append_2#(append_2(l1,l2),l3) -> append_2#(l1,append_2(l2,l3)) and R consists of: r1: app(append_0(),y) -> append_1(y) r2: app(append_1(x1),y) -> append_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: append_2(nil_0(),l) -> l r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l)) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3)) r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2)) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: append_2#(cons_2(h,t),l) -> append_2#(t,l) p2: append_2#(append_2(l1,l2),l3) -> append_2#(l1,append_2(l2,l3)) and R consists of: r1: app(append_0(),y) -> append_1(y) r2: app(append_1(x1),y) -> append_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: append_2(nil_0(),l) -> l r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l)) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3)) r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: model-based reduction triple (=A, >=A!, >A!): finite algebra: carrier: {0,1} interpretations: append_2#_A(0,0) = 0 append_2#_A(0,1) = 0 append_2#_A(1,0) = 0 append_2#_A(1,1) = 0 cons_2_A(0,0) = 0 cons_2_A(0,1) = 1 cons_2_A(1,0) = 0 cons_2_A(1,1) = 1 append_2_A(0,0) = 0 append_2_A(0,1) = 1 append_2_A(1,0) = 0 append_2_A(1,1) = 1 app_A(0,0) = 0 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 0 append_0_A = 1 append_1_A(0) = 0 append_1_A(1) = 0 cons_0_A = 1 cons_1_A(0) = 0 cons_1_A(1) = 0 map_0_A = 0 map_1_A(0) = 0 map_1_A(1) = 1 map_2_A(0,0) = 0 map_2_A(0,1) = 1 map_2_A(1,0) = 0 map_2_A(1,1) = 0 nil_0_A = 0 append_2#!_A(0,0) = 0 append_2#!_A(0,1) = 0 append_2#!_A(1,0) = 0 append_2#!_A(1,1) = 0 cons_2!_A(0,0) = 0 cons_2!_A(0,1) = 0 cons_2!_A(1,0) = 0 cons_2!_A(1,1) = 0 append_2!_A(0,0) = 0 append_2!_A(0,1) = 0 append_2!_A(1,0) = 0 append_2!_A(1,1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 append_0!_A = 1 append_1!_A(0) = 0 append_1!_A(1) = 0 cons_0!_A = 0 cons_1!_A(0) = 0 cons_1!_A(1) = 0 map_0!_A = 1 map_1!_A(0) = 0 map_1!_A(1) = 0 map_2!_A(0,0) = 1 map_2!_A(0,1) = 1 map_2!_A(1,0) = 1 map_2!_A(1,1) = 1 nil_0!_A = 1 partial status: pi(append_2#) = [1] pi(cons_2) = [1, 2] pi(append_2) = [1, 2] pi(app) = [1, 2] pi(append_0) = [] pi(append_1) = [1] pi(cons_0) = [] pi(cons_1) = [1] pi(map_0) = [] pi(map_1) = [1] pi(map_2) = [1, 2] pi(nil_0) = [] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: append_2#(cons_2(h,t),l) -> append_2#(t,l) and R consists of: r1: app(append_0(),y) -> append_1(y) r2: app(append_1(x1),y) -> append_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: append_2(nil_0(),l) -> l r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l)) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3)) r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2)) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: append_2#(cons_2(h,t),l) -> append_2#(t,l) and R consists of: r1: app(append_0(),y) -> append_1(y) r2: app(append_1(x1),y) -> append_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: append_2(nil_0(),l) -> l r8: append_2(cons_2(h,t),l) -> cons_2(h,append_2(t,l)) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(h,t)) -> cons_2(app(f,h),map_2(f,t)) r11: append_2(append_2(l1,l2),l3) -> append_2(l1,append_2(l2,l3)) r12: map_2(f,append_2(l1,l2)) -> append_2(map_2(f,l1),map_2(f,l2)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: model-based reduction triple (=A, >=A!, >A!): finite algebra: carrier: {0,1} interpretations: append_2#_A(0,0) = 0 append_2#_A(0,1) = 0 append_2#_A(1,0) = 0 append_2#_A(1,1) = 0 cons_2_A(0,0) = 0 cons_2_A(0,1) = 1 cons_2_A(1,0) = 0 cons_2_A(1,1) = 1 app_A(0,0) = 1 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 1 append_0_A = 0 append_1_A(0) = 1 append_1_A(1) = 1 append_2_A(0,0) = 0 append_2_A(0,1) = 1 append_2_A(1,0) = 0 append_2_A(1,1) = 1 cons_0_A = 0 cons_1_A(0) = 1 cons_1_A(1) = 1 map_0_A = 0 map_1_A(0) = 1 map_1_A(1) = 1 map_2_A(0,0) = 0 map_2_A(0,1) = 1 map_2_A(1,0) = 0 map_2_A(1,1) = 1 nil_0_A = 1 append_2#!_A(0,0) = 0 append_2#!_A(0,1) = 0 append_2#!_A(1,0) = 1 append_2#!_A(1,1) = 0 cons_2!_A(0,0) = 0 cons_2!_A(0,1) = 0 cons_2!_A(1,0) = 0 cons_2!_A(1,1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 append_0!_A = 0 append_1!_A(0) = 0 append_1!_A(1) = 0 append_2!_A(0,0) = 0 append_2!_A(0,1) = 0 append_2!_A(1,0) = 0 append_2!_A(1,1) = 0 cons_0!_A = 1 cons_1!_A(0) = 0 cons_1!_A(1) = 0 map_0!_A = 0 map_1!_A(0) = 0 map_1!_A(1) = 0 map_2!_A(0,0) = 1 map_2!_A(0,1) = 1 map_2!_A(1,0) = 1 map_2!_A(1,1) = 1 nil_0!_A = 1 partial status: pi(append_2#) = [1, 2] pi(cons_2) = [1, 2] pi(app) = [1, 2] pi(append_0) = [] pi(append_1) = [1] pi(append_2) = [1, 2] pi(cons_0) = [] pi(cons_1) = [1] pi(map_0) = [] pi(map_1) = [1] pi(map_2) = [1, 2] pi(nil_0) = [] The following rules are removed: p1 Then no dependency pair remains.