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(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)) inc() -> app(map(),app(plus(),app(s(),|0|()))) 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(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)) map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) inc_0() -> map_1(plus_1(s_1(|0|_0()))) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(map_1(x1),y) -> map_2#(x1,y) p2: app#(plus_1(x1),y) -> plus_2#(x1,y) p3: plus_2#(s_1(x),y) -> plus_2#(x,y) p4: map_2#(f,cons_2(x,xs)) -> app#(f,x) p5: 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(map_0(),y) -> map_1(y) r4: app(map_1(x1),y) -> map_2(x1,y) r5: app(plus_0(),y) -> plus_1(y) r6: app(plus_1(x1),y) -> plus_2(x1,y) r7: app(s_0(),y) -> s_1(y) r8: plus_2(|0|_0(),y) -> y r9: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r10: map_2(f,nil_0()) -> nil_0() r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r12: inc_0() -> map_1(plus_1(s_1(|0|_0()))) The estimated dependency graph contains the following SCCs: {p1, p4, p5} {p3} -- Reduction pair. 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(x,xs)) -> map_2#(f,xs) p3: 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(map_0(),y) -> map_1(y) r4: app(map_1(x1),y) -> map_2(x1,y) r5: app(plus_0(),y) -> plus_1(y) r6: app(plus_1(x1),y) -> plus_2(x1,y) r7: app(s_0(),y) -> s_1(y) r8: plus_2(|0|_0(),y) -> y r9: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r10: map_2(f,nil_0()) -> nil_0() r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r12: inc_0() -> map_1(plus_1(s_1(|0|_0()))) The set of usable rules consists of r1, r3, r5, r7, r12 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: app#_A(x1,x2) = x2 + 9 map_1_A(x1) = max{2, x1 - 12} map_2#_A(x1,x2) = max{8, x2 + 3} cons_2_A(x1,x2) = max{11, x1 + 10, x2 + 6} app_A(x1,x2) = max{5, x1 - 18, x2 - 11} cons_0_A = 3 cons_1_A(x1) = max{4, x1 - 12} map_0_A = 1 plus_0_A = 1 plus_1_A(x1) = max{2, x1 - 12} s_0_A = 1 s_1_A(x1) = max{2, x1 - 32} inc_0_A = 3 |0|_0_A = 0 base order 2: precedence: app > cons_2 = inc_0 > map_2# = cons_1 = plus_1 = s_1 = |0|_0 > app# = map_1 = cons_0 = plus_0 = s_0 > map_0 partial status: pi(app#) = [] pi(map_1) = [] pi(map_2#) = [2] pi(cons_2) = [2] pi(app) = [] pi(cons_0) = [] pi(cons_1) = [] pi(map_0) = [] pi(plus_0) = [] pi(plus_1) = [] pi(s_0) = [] pi(s_1) = [] pi(inc_0) = [] pi(|0|_0) = [] The following rules are removed: p1 -- 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(map_0(),y) -> map_1(y) r4: app(map_1(x1),y) -> map_2(x1,y) r5: app(plus_0(),y) -> plus_1(y) r6: app(plus_1(x1),y) -> plus_2(x1,y) r7: app(s_0(),y) -> s_1(y) r8: plus_2(|0|_0(),y) -> y r9: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r10: map_2(f,nil_0()) -> nil_0() r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r12: inc_0() -> map_1(plus_1(s_1(|0|_0()))) 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(map_0(),y) -> map_1(y) r4: app(map_1(x1),y) -> map_2(x1,y) r5: app(plus_0(),y) -> plus_1(y) r6: app(plus_1(x1),y) -> plus_2(x1,y) r7: app(s_0(),y) -> s_1(y) r8: plus_2(|0|_0(),y) -> y r9: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r10: map_2(f,nil_0()) -> nil_0() r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r12: inc_0() -> map_1(plus_1(s_1(|0|_0()))) The set of usable rules consists of r1, r3, r5, r7, r12 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: map_2#_A(x1,x2) = max{4, x2 - 4} cons_2_A(x1,x2) = max{x1 - 12, x2 + 9} app_A(x1,x2) = max{11, x1 - 3, x2 - 1} cons_0_A = 1 cons_1_A(x1) = max{2, x1 - 13} map_0_A = 1 map_1_A(x1) = max{2, x1 - 12} plus_0_A = 1 plus_1_A(x1) = max{2, x1 - 3} s_0_A = 1 s_1_A(x1) = max{2, x1 - 8} inc_0_A = 3 |0|_0_A = 2 base order 2: precedence: app > inc_0 > cons_2 = map_1 = plus_1 = s_1 = |0|_0 > map_2# = cons_1 = map_0 = plus_0 = s_0 > cons_0 partial status: pi(map_2#) = [] pi(cons_2) = [2] pi(app) = [] pi(cons_0) = [] pi(cons_1) = [] pi(map_0) = [] pi(map_1) = [] pi(plus_0) = [] pi(plus_1) = [] pi(s_0) = [] pi(s_1) = [] pi(inc_0) = [] pi(|0|_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(map_0(),y) -> map_1(y) r4: app(map_1(x1),y) -> map_2(x1,y) r5: app(plus_0(),y) -> plus_1(y) r6: app(plus_1(x1),y) -> plus_2(x1,y) r7: app(s_0(),y) -> s_1(y) r8: plus_2(|0|_0(),y) -> y r9: plus_2(s_1(x),y) -> s_1(plus_2(x,y)) r10: map_2(f,nil_0()) -> nil_0() r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r12: inc_0() -> map_1(plus_1(s_1(|0|_0()))) The set of usable rules consists of r1, r2, r3, r5, r12 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: plus_2#_A(x1,x2) = max{5, x1 + 1} s_1_A(x1) = x1 + 7 app_A(x1,x2) = max{3, x1 - 4, x2 - 5} cons_0_A = 1 cons_1_A(x1) = max{2, x1 - 6} cons_2_A(x1,x2) = max{3, x1 - 11, x2 - 9} map_0_A = 1 map_1_A(x1) = max{2, x1 - 16} plus_0_A = 2 plus_1_A(x1) = max{3, x1 - 8} inc_0_A = 8 |0|_0_A = 0 base order 2: precedence: inc_0 > s_1 = |0|_0 > app = map_0 > plus_2# = cons_2 = map_1 = plus_1 > cons_1 = plus_0 > cons_0 partial status: pi(plus_2#) = [1] pi(s_1) = [1] pi(app) = [] pi(cons_0) = [] pi(cons_1) = [] pi(cons_2) = [] pi(map_0) = [] pi(map_1) = [] pi(plus_0) = [] pi(plus_1) = [] pi(inc_0) = [] pi(|0|_0) = [] The following rules are removed: p1 Then no dependency pair remains.