YES We show the termination of the TRS R: 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)) app(app(minus(),x),|0|()) -> x app(app(minus(),app(s(),x)),app(s(),y)) -> app(app(minus(),app(p(),app(s(),x))),app(p(),app(s(),y))) app(p(),app(s(),x)) -> x app(app(div(),|0|()),app(s(),y)) -> |0|() app(app(div(),app(s(),x)),app(s(),y)) -> app(s(),app(app(div(),app(app(minus(),x),y)),app(s(),y))) 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(div_0(),y) -> div_1(y) app(div_1(x1),y) -> div_2(x1,y) app(map_0(),y) -> map_1(y) app(map_1(x1),y) -> map_2(x1,y) app(minus_0(),y) -> minus_1(y) app(minus_1(x1),y) -> minus_2(x1,y) app(p_0(),y) -> p_1(y) app(s_0(),y) -> s_1(y) map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) minus_2(x,|0|_0()) -> x minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y))) p_1(s_1(x)) -> x div_2(|0|_0(),s_1(y)) -> |0|_0() div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y))) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(div_1(x1),y) -> div_2#(x1,y) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: app#(minus_1(x1),y) -> minus_2#(x1,y) p4: app#(p_0(),y) -> p_1#(y) p5: map_2#(f,cons_2(x,xs)) -> app#(f,x) p6: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p7: minus_2#(s_1(x),s_1(y)) -> minus_2#(p_1(s_1(x)),p_1(s_1(y))) p8: minus_2#(s_1(x),s_1(y)) -> p_1#(s_1(x)) p9: minus_2#(s_1(x),s_1(y)) -> p_1#(s_1(y)) p10: div_2#(s_1(x),s_1(y)) -> div_2#(minus_2(x,y),s_1(y)) p11: div_2#(s_1(x),s_1(y)) -> minus_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(div_0(),y) -> div_1(y) r4: app(div_1(x1),y) -> div_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(minus_0(),y) -> minus_1(y) r8: app(minus_1(x1),y) -> minus_2(x1,y) r9: app(p_0(),y) -> p_1(y) r10: app(s_0(),y) -> s_1(y) r11: map_2(f,nil_0()) -> nil_0() r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r13: minus_2(x,|0|_0()) -> x r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y))) r15: p_1(s_1(x)) -> x r16: div_2(|0|_0(),s_1(y)) -> |0|_0() r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y))) The estimated dependency graph contains the following SCCs: {p2, p5, p6} {p10} {p7} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> app#(f,x) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: 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(div_0(),y) -> div_1(y) r4: app(div_1(x1),y) -> div_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(minus_0(),y) -> minus_1(y) r8: app(minus_1(x1),y) -> minus_2(x1,y) r9: app(p_0(),y) -> p_1(y) r10: app(s_0(),y) -> s_1(y) r11: map_2(f,nil_0()) -> nil_0() r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r13: minus_2(x,|0|_0()) -> x r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y))) r15: p_1(s_1(x)) -> x r16: div_2(|0|_0(),s_1(y)) -> |0|_0() r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y))) The set of usable rules consists of r1, r3, r5, r7, r10, r16 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{3, x2 + 1} cons_2_A(x1,x2) = max{x1 + 5, x2 + 5} app#_A(x1,x2) = x2 + 4 map_1_A(x1) = max{2, x1 - 12} app_A(x1,x2) = max{18, x1 - 6, x2 - 11} cons_0_A = 3 cons_1_A(x1) = max{4, x1 - 12} div_0_A = 4 div_1_A(x1) = max{5, x1 - 12} map_0_A = 1 minus_0_A = 5 minus_1_A(x1) = max{4, x1 - 12} s_0_A = 5 s_1_A(x1) = max{6, x1 - 39} div_2_A(x1,x2) = max{17, x1 - 19, x2 - 29} |0|_0_A = 5 base order 2: precedence: app > cons_2 = map_1 = s_1 = div_2 > map_2# = cons_1 = div_1 = map_0 = minus_1 = s_0 = |0|_0 > app# = cons_0 = div_0 = minus_0 partial status: pi(map_2#) = [2] pi(cons_2) = [2] pi(app#) = [] pi(map_1) = [] pi(app) = [] pi(cons_0) = [] pi(cons_1) = [] pi(div_0) = [] pi(div_1) = [] pi(map_0) = [] pi(minus_0) = [] pi(minus_1) = [] pi(s_0) = [] pi(s_1) = [] pi(div_2) = [] pi(|0|_0) = [] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: map_2#(f,cons_2(x,xs)) -> app#(f,x) p2: 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(div_0(),y) -> div_1(y) r4: app(div_1(x1),y) -> div_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(minus_0(),y) -> minus_1(y) r8: app(minus_1(x1),y) -> minus_2(x1,y) r9: app(p_0(),y) -> p_1(y) r10: app(s_0(),y) -> s_1(y) r11: map_2(f,nil_0()) -> nil_0() r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r13: minus_2(x,|0|_0()) -> x r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y))) r15: p_1(s_1(x)) -> x r16: div_2(|0|_0(),s_1(y)) -> |0|_0() r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y))) 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(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(div_0(),y) -> div_1(y) r4: app(div_1(x1),y) -> div_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(minus_0(),y) -> minus_1(y) r8: app(minus_1(x1),y) -> minus_2(x1,y) r9: app(p_0(),y) -> p_1(y) r10: app(s_0(),y) -> s_1(y) r11: map_2(f,nil_0()) -> nil_0() r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r13: minus_2(x,|0|_0()) -> x r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y))) r15: p_1(s_1(x)) -> x r16: div_2(|0|_0(),s_1(y)) -> |0|_0() r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y))) The set of usable rules consists of r1, r3, r5, r7, r10, r13, r16 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{x1 + 8, x2 + 16} cons_2_A(x1,x2) = max{6, x1 - 8, x2 + 1} app_A(x1,x2) = max{7, x1 - 9, x2 - 20} cons_0_A = 4 cons_1_A(x1) = max{5, x1 - 20} div_0_A = 1 div_1_A(x1) = max{2, x1 - 21} map_0_A = 18 map_1_A(x1) = max{8, x1 - 21} minus_0_A = 2 minus_1_A(x1) = max{3, x1 - 21} s_0_A = 8 s_1_A(x1) = max{6, x1 - 32} minus_2_A(x1,x2) = max{4, x1, x2} |0|_0_A = 0 div_2_A(x1,x2) = max{3, x1 - 31, x2 - 32} base order 2: precedence: map_2# = minus_2 > cons_2 = app = map_0 > cons_1 = map_1 = minus_1 = s_1 = div_2 > cons_0 = div_1 = minus_0 = s_0 = |0|_0 > div_0 partial status: pi(map_2#) = [2] pi(cons_2) = [2] pi(app) = [] pi(cons_0) = [] pi(cons_1) = [] pi(div_0) = [] pi(div_1) = [] pi(map_0) = [] pi(map_1) = [] pi(minus_0) = [] pi(minus_1) = [] pi(s_0) = [] pi(s_1) = [] pi(minus_2) = [1, 2] pi(|0|_0) = [] pi(div_2) = [] 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: div_2#(s_1(x),s_1(y)) -> div_2#(minus_2(x,y),s_1(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(div_0(),y) -> div_1(y) r4: app(div_1(x1),y) -> div_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(minus_0(),y) -> minus_1(y) r8: app(minus_1(x1),y) -> minus_2(x1,y) r9: app(p_0(),y) -> p_1(y) r10: app(s_0(),y) -> s_1(y) r11: map_2(f,nil_0()) -> nil_0() r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r13: minus_2(x,|0|_0()) -> x r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y))) r15: p_1(s_1(x)) -> x r16: div_2(|0|_0(),s_1(y)) -> |0|_0() r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y))) The set of usable rules consists of r1, r2, r3, r5, r7, r9, r13, r14, r15, r16 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: div_2#_A(x1,x2) = max{1, x1 - 10} s_1_A(x1) = x1 + 12 minus_2_A(x1,x2) = max{3, x1 + 1} app_A(x1,x2) = max{5, x1 - 7, x2 - 6} cons_0_A = 6 cons_1_A(x1) = max{5, x1 - 7} cons_2_A(x1,x2) = max{4, x1 - 15, x2 - 27} div_0_A = 2 div_1_A(x1) = max{3, x1 - 7} map_0_A = 6 map_1_A(x1) = max{4, x1 - 23} minus_0_A = 1 minus_1_A(x1) = max{2, x1 - 6} p_0_A = 12 p_1_A(x1) = max{4, x1 - 7} |0|_0_A = 1 div_2_A(x1,x2) = 2 base order 2: precedence: div_2# = s_1 = minus_2 = div_1 = p_1 > app = div_0 = minus_0 = p_0 = |0|_0 > cons_2 = map_1 = minus_1 = div_2 > cons_1 = map_0 > cons_0 partial status: pi(div_2#) = [] pi(s_1) = [1] pi(minus_2) = [] pi(app) = [] pi(cons_0) = [] pi(cons_1) = [] pi(cons_2) = [] pi(div_0) = [] pi(div_1) = [] pi(map_0) = [] pi(map_1) = [] pi(minus_0) = [] pi(minus_1) = [] pi(p_0) = [] pi(p_1) = [] pi(|0|_0) = [] pi(div_2) = [] 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: minus_2#(s_1(x),s_1(y)) -> minus_2#(p_1(s_1(x)),p_1(s_1(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(div_0(),y) -> div_1(y) r4: app(div_1(x1),y) -> div_2(x1,y) r5: app(map_0(),y) -> map_1(y) r6: app(map_1(x1),y) -> map_2(x1,y) r7: app(minus_0(),y) -> minus_1(y) r8: app(minus_1(x1),y) -> minus_2(x1,y) r9: app(p_0(),y) -> p_1(y) r10: app(s_0(),y) -> s_1(y) r11: map_2(f,nil_0()) -> nil_0() r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r13: minus_2(x,|0|_0()) -> x r14: minus_2(s_1(x),s_1(y)) -> minus_2(p_1(s_1(x)),p_1(s_1(y))) r15: p_1(s_1(x)) -> x r16: div_2(|0|_0(),s_1(y)) -> |0|_0() r17: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y))) The set of usable rules consists of r1, r2, r3, r5, r7, r15 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: minus_2#_A(x1,x2) = max{1, x1} s_1_A(x1) = x1 + 16 p_1_A(x1) = max{2, x1 - 15} app_A(x1,x2) = max{5, x1 - 6, x2 - 21} cons_0_A = 5 cons_1_A(x1) = max{2, x1 - 22} cons_2_A(x1,x2) = max{3, x1 - 28, x2 - 33} div_0_A = 2 div_1_A(x1) = max{3, x1 - 25} map_0_A = 2 map_1_A(x1) = max{3, x1 - 35} minus_0_A = 4 minus_1_A(x1) = max{5, x1 - 21} base order 2: precedence: minus_2# = s_1 > app = cons_0 > p_1 = cons_2 = div_1 = map_1 = minus_1 > cons_1 = div_0 = map_0 = minus_0 partial status: pi(minus_2#) = [1] pi(s_1) = [1] pi(p_1) = [] pi(app) = [] pi(cons_0) = [] pi(cons_1) = [] pi(cons_2) = [] pi(div_0) = [] pi(div_1) = [] pi(map_0) = [] pi(map_1) = [] pi(minus_0) = [] pi(minus_1) = [] The following rules are removed: p1 Then no dependency pair remains.