YES We show the termination of the TRS R: app(app(plus(),|0|()),y) -> y app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y)) app(inc(),xs) -> app(app(map(),app(plus(),app(s(),|0|()))),xs) app(app(map(),f),nil()) -> nil() app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs)) Instead we prove termination of the uncurried system: app(cons_0(),y) -> cons_1(y) app(cons_1(x1),y) -> cons_2(x1,y) app(inc_0(),y) -> inc_1(y) app(map_0(),y) -> map_1(y) app(map_1(x1),y) -> map_2(x1,y) app(plus_0(),y) -> plus_1(y) app(plus_1(x1),y) -> plus_2(x1,y) app(s_0(),y) -> s_1(y) plus_2(|0|_0(),y) -> y plus_2(s_1(x),y) -> s_1(plus_2(x,y)) inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(inc_0(),y) -> inc_1#(y) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: app#(plus_1(x1),y) -> plus_2#(x1,y) p4: plus_2#(s_1(x),y) -> plus_2#(x,y) p5: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs) p6: map_2#(f,cons_2(x,xs)) -> app#(f,x) p7: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(inc_0(),y) -> inc_1(y) r4: app(map_0(),y) -> map_1(y) r5: app(map_1(x1),y) -> map_2(x1,y) r6: app(plus_0(),y) -> plus_1(y) r7: app(plus_1(x1),y) -> plus_2(x1,y) r8: app(s_0(),y) -> s_1(y) r9: plus_2(|0|_0(),y) -> y r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r12: map_2(f,nil_0()) -> nil_0() r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The estimated dependency graph contains the following SCCs: {p1, p2, p5, p6, p7} {p4} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(inc_0(),y) -> inc_1#(y) p2: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs) p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p4: map_2#(f,cons_2(x,xs)) -> app#(f,x) p5: app#(map_1(x1),y) -> map_2#(x1,y) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(inc_0(),y) -> inc_1(y) r4: app(map_0(),y) -> map_1(y) r5: app(map_1(x1),y) -> map_2(x1,y) r6: app(plus_0(),y) -> plus_1(y) r7: app(plus_1(x1),y) -> plus_2(x1,y) r8: app(s_0(),y) -> s_1(y) r9: plus_2(|0|_0(),y) -> y r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r12: map_2(f,nil_0()) -> nil_0() r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13 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: app#_A(0,0) = 0 app#_A(0,1) = 0 app#_A(1,0) = 0 app#_A(1,1) = 0 inc_0_A = 0 inc_1#_A(0) = 0 inc_1#_A(1) = 0 map_2#_A(0,0) = 0 map_2#_A(0,1) = 0 map_2#_A(1,0) = 0 map_2#_A(1,1) = 0 plus_1_A(0) = 0 plus_1_A(1) = 1 s_1_A(0) = 0 s_1_A(1) = 1 |0|_0_A = 1 cons_2_A(0,0) = 0 cons_2_A(0,1) = 1 cons_2_A(1,0) = 0 cons_2_A(1,1) = 1 map_1_A(0) = 0 map_1_A(1) = 1 app_A(0,0) = 0 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 1 cons_0_A = 1 cons_1_A(0) = 0 cons_1_A(1) = 1 inc_1_A(0) = 0 inc_1_A(1) = 1 map_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 plus_0_A = 0 plus_2_A(0,0) = 0 plus_2_A(0,1) = 1 plus_2_A(1,0) = 0 plus_2_A(1,1) = 1 s_0_A = 0 nil_0_A = 0 app#!_A(0,0) = 1 app#!_A(0,1) = 1 app#!_A(1,0) = 0 app#!_A(1,1) = 0 inc_0!_A = 0 inc_1#!_A(0) = 0 inc_1#!_A(1) = 0 map_2#!_A(0,0) = 1 map_2#!_A(0,1) = 1 map_2#!_A(1,0) = 0 map_2#!_A(1,1) = 0 plus_1!_A(0) = 0 plus_1!_A(1) = 0 s_1!_A(0) = 0 s_1!_A(1) = 0 |0|_0!_A = 0 cons_2!_A(0,0) = 1 cons_2!_A(0,1) = 1 cons_2!_A(1,0) = 1 cons_2!_A(1,1) = 1 map_1!_A(0) = 1 map_1!_A(1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 cons_0!_A = 0 cons_1!_A(0) = 1 cons_1!_A(1) = 0 inc_1!_A(0) = 1 inc_1!_A(1) = 1 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 plus_0!_A = 1 plus_2!_A(0,0) = 0 plus_2!_A(0,1) = 0 plus_2!_A(1,0) = 0 plus_2!_A(1,1) = 0 s_0!_A = 1 nil_0!_A = 0 partial status: pi(app#) = [] pi(inc_0) = [] pi(inc_1#) = [] pi(map_2#) = [] pi(plus_1) = [] pi(s_1) = [] pi(|0|_0) = [] pi(cons_2) = [2] pi(map_1) = [1] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [] pi(inc_1) = [1] pi(map_0) = [] pi(map_2) = [2] pi(plus_0) = [] pi(plus_2) = [2] pi(s_0) = [] pi(nil_0) = [] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: inc_1#(xs) -> map_2#(plus_1(s_1(|0|_0())),xs) p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p3: map_2#(f,cons_2(x,xs)) -> app#(f,x) p4: app#(map_1(x1),y) -> map_2#(x1,y) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(inc_0(),y) -> inc_1(y) r4: app(map_0(),y) -> map_1(y) r5: app(map_1(x1),y) -> map_2(x1,y) r6: app(plus_0(),y) -> plus_1(y) r7: app(plus_1(x1),y) -> plus_2(x1,y) r8: app(s_0(),y) -> s_1(y) r9: plus_2(|0|_0(),y) -> y r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r12: map_2(f,nil_0()) -> nil_0() r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The estimated dependency graph contains the following SCCs: {p2, p3, p4} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p2: map_2#(f,cons_2(x,xs)) -> app#(f,x) p3: app#(map_1(x1),y) -> map_2#(x1,y) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(inc_0(),y) -> inc_1(y) r4: app(map_0(),y) -> map_1(y) r5: app(map_1(x1),y) -> map_2(x1,y) r6: app(plus_0(),y) -> plus_1(y) r7: app(plus_1(x1),y) -> plus_2(x1,y) r8: app(s_0(),y) -> s_1(y) r9: plus_2(|0|_0(),y) -> y r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r12: map_2(f,nil_0()) -> nil_0() r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13 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) = 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) = 1 app_A(0,0) = 0 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 0 cons_0_A = 0 cons_1_A(0) = 0 cons_1_A(1) = 1 inc_0_A = 0 inc_1_A(0) = 0 inc_1_A(1) = 1 map_0_A = 0 map_2_A(0,0) = 0 map_2_A(0,1) = 1 map_2_A(1,0) = 0 map_2_A(1,1) = 0 plus_0_A = 0 plus_1_A(0) = 0 plus_1_A(1) = 1 plus_2_A(0,0) = 0 plus_2_A(0,1) = 1 plus_2_A(1,0) = 0 plus_2_A(1,1) = 0 s_0_A = 0 s_1_A(0) = 0 s_1_A(1) = 1 |0|_0_A = 0 nil_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 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 map_1!_A(0) = 1 map_1!_A(1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 cons_0!_A = 1 cons_1!_A(0) = 0 cons_1!_A(1) = 0 inc_0!_A = 1 inc_1!_A(0) = 1 inc_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) = 1 map_2!_A(1,1) = 1 plus_0!_A = 1 plus_1!_A(0) = 1 plus_1!_A(1) = 1 plus_2!_A(0,0) = 0 plus_2!_A(0,1) = 1 plus_2!_A(1,0) = 0 plus_2!_A(1,1) = 0 s_0!_A = 0 s_1!_A(0) = 0 s_1!_A(1) = 0 |0|_0!_A = 0 nil_0!_A = 1 partial status: pi(map_2#) = [1] pi(cons_2) = [] pi(app#) = [1] pi(map_1) = [1] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [1] pi(inc_0) = [] pi(inc_1) = [] pi(map_0) = [] pi(map_2) = [] pi(plus_0) = [] pi(plus_1) = [] pi(plus_2) = [2] pi(s_0) = [] pi(s_1) = [] pi(|0|_0) = [] pi(nil_0) = [] The following rules are removed: p3 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p2: map_2#(f,cons_2(x,xs)) -> app#(f,x) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(inc_0(),y) -> inc_1(y) r4: app(map_0(),y) -> map_1(y) r5: app(map_1(x1),y) -> map_2(x1,y) r6: app(plus_0(),y) -> plus_1(y) r7: app(plus_1(x1),y) -> plus_2(x1,y) r8: app(s_0(),y) -> s_1(y) r9: plus_2(|0|_0(),y) -> y r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r12: map_2(f,nil_0()) -> nil_0() r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(inc_0(),y) -> inc_1(y) r4: app(map_0(),y) -> map_1(y) r5: app(map_1(x1),y) -> map_2(x1,y) r6: app(plus_0(),y) -> plus_1(y) r7: app(plus_1(x1),y) -> plus_2(x1,y) r8: app(s_0(),y) -> s_1(y) r9: plus_2(|0|_0(),y) -> y r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r12: map_2(f,nil_0()) -> nil_0() r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13 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) = 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) = 1 cons_0_A = 0 cons_1_A(0) = 0 cons_1_A(1) = 0 inc_0_A = 0 inc_1_A(0) = 0 inc_1_A(1) = 0 map_0_A = 1 map_1_A(0) = 0 map_1_A(1) = 1 map_2_A(0,0) = 0 map_2_A(0,1) = 0 map_2_A(1,0) = 0 map_2_A(1,1) = 1 plus_0_A = 1 plus_1_A(0) = 0 plus_1_A(1) = 1 plus_2_A(0,0) = 0 plus_2_A(0,1) = 0 plus_2_A(1,0) = 0 plus_2_A(1,1) = 1 s_0_A = 0 s_1_A(0) = 0 s_1_A(1) = 0 |0|_0_A = 1 nil_0_A = 0 map_2#!_A(0,0) = 1 map_2#!_A(0,1) = 0 map_2#!_A(1,0) = 1 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) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 cons_0!_A = 1 cons_1!_A(0) = 0 cons_1!_A(1) = 0 inc_0!_A = 0 inc_1!_A(0) = 0 inc_1!_A(1) = 0 map_0!_A = 1 map_1!_A(0) = 1 map_1!_A(1) = 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 plus_0!_A = 0 plus_1!_A(0) = 0 plus_1!_A(1) = 0 plus_2!_A(0,0) = 1 plus_2!_A(0,1) = 1 plus_2!_A(1,0) = 0 plus_2!_A(1,1) = 1 s_0!_A = 1 s_1!_A(0) = 1 s_1!_A(1) = 0 |0|_0!_A = 0 nil_0!_A = 0 partial status: pi(map_2#) = [1, 2] pi(cons_2) = [2] pi(app) = [1, 2] pi(cons_0) = [] pi(cons_1) = [1] pi(inc_0) = [] pi(inc_1) = [1] pi(map_0) = [] pi(map_1) = [] pi(map_2) = [2] pi(plus_0) = [] pi(plus_1) = [1] pi(plus_2) = [1, 2] pi(s_0) = [] pi(s_1) = [] pi(|0|_0) = [] 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: plus_2#(s_1(x),y) -> plus_2#(x,y) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(inc_0(),y) -> inc_1(y) r4: app(map_0(),y) -> map_1(y) r5: app(map_1(x1),y) -> map_2(x1,y) r6: app(plus_0(),y) -> plus_1(y) r7: app(plus_1(x1),y) -> plus_2(x1,y) r8: app(s_0(),y) -> s_1(y) r9: plus_2(|0|_0(),y) -> y r10: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r11: inc_1(xs) -> map_2(plus_1(s_1(|0|_0())),xs) r12: map_2(f,nil_0()) -> nil_0() r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13 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: plus_2#_A(0,0) = 0 plus_2#_A(0,1) = 0 plus_2#_A(1,0) = 0 plus_2#_A(1,1) = 0 s_1_A(0) = 0 s_1_A(1) = 1 app_A(0,0) = 0 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 0 cons_0_A = 0 cons_1_A(0) = 0 cons_1_A(1) = 1 cons_2_A(0,0) = 0 cons_2_A(0,1) = 1 cons_2_A(1,0) = 0 cons_2_A(1,1) = 0 inc_0_A = 0 inc_1_A(0) = 0 inc_1_A(1) = 1 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 plus_0_A = 0 plus_1_A(0) = 0 plus_1_A(1) = 1 plus_2_A(0,0) = 0 plus_2_A(0,1) = 1 plus_2_A(1,0) = 0 plus_2_A(1,1) = 0 s_0_A = 0 |0|_0_A = 0 nil_0_A = 0 plus_2#!_A(0,0) = 0 plus_2#!_A(0,1) = 0 plus_2#!_A(1,0) = 0 plus_2#!_A(1,1) = 0 s_1!_A(0) = 0 s_1!_A(1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 cons_0!_A = 1 cons_1!_A(0) = 1 cons_1!_A(1) = 1 cons_2!_A(0,0) = 1 cons_2!_A(0,1) = 1 cons_2!_A(1,0) = 1 cons_2!_A(1,1) = 1 inc_0!_A = 0 inc_1!_A(0) = 1 inc_1!_A(1) = 1 map_0!_A = 1 map_1!_A(0) = 0 map_1!_A(1) = 1 map_2!_A(0,0) = 1 map_2!_A(0,1) = 1 map_2!_A(1,0) = 1 map_2!_A(1,1) = 1 plus_0!_A = 1 plus_1!_A(0) = 0 plus_1!_A(1) = 0 plus_2!_A(0,0) = 1 plus_2!_A(0,1) = 1 plus_2!_A(1,0) = 1 plus_2!_A(1,1) = 1 s_0!_A = 0 |0|_0!_A = 0 nil_0!_A = 1 partial status: pi(plus_2#) = [1] pi(s_1) = [1] pi(app) = [1, 2] pi(cons_0) = [] pi(cons_1) = [] pi(cons_2) = [] pi(inc_0) = [] pi(inc_1) = [] pi(map_0) = [] pi(map_1) = [] pi(map_2) = [] pi(plus_0) = [] pi(plus_1) = [1] pi(plus_2) = [1, 2] pi(s_0) = [] pi(|0|_0) = [] pi(nil_0) = [] The following rules are removed: p1 Then no dependency pair remains.