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(app(comp(),f),g),x) -> app(f,app(g,x)) app(twice(),f) -> app(app(comp(),f),f) Instead we prove termination of the uncurried system: app(comp_0(),y) -> comp_1(y) app(comp_1(x1),y) -> comp_2(x1,y) app(comp_2(x1,x2),y) -> comp_3(x1,x2,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) app(twice_0(),y) -> twice_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)) comp_3(f,g,x) -> app(f,app(g,x)) twice_1(f) -> comp_2(f,f) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y) p2: app#(map_1(x1),y) -> map_2#(x1,y) p3: app#(twice_0(),y) -> twice_1#(y) p4: map_2#(f,cons_2(x,xs)) -> app#(f,x) p5: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p6: comp_3#(f,g,x) -> app#(f,app(g,x)) p7: comp_3#(f,g,x) -> app#(g,x) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The estimated dependency graph contains the following SCCs: {p1, p2, p4, p5, p6, p7} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y) p2: comp_3#(f,g,x) -> app#(g,x) p3: app#(map_1(x1),y) -> map_2#(x1,y) p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p5: map_2#(f,cons_2(x,xs)) -> app#(f,x) p6: comp_3#(f,g,x) -> app#(f,app(g,x)) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The set of usable rules consists of r1, r4, r5 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: app#_A(x1,x2) = max{4, x1 - 1} comp_2_A(x1,x2) = max{10, x1 + 7, x2 + 7} comp_3#_A(x1,x2,x3) = max{x1 + 5, x2 + 5} map_1_A(x1) = x1 + 7 map_2#_A(x1,x2) = x1 + 5 cons_2_A(x1,x2) = max{3, x1 - 24, x2 - 24} app_A(x1,x2) = max{9, x1 - 11, x2 - 11} comp_0_A = 7 comp_1_A(x1) = max{8, x1 - 12} cons_0_A = 1 cons_1_A(x1) = max{2, x1 - 12} base order 2: precedence: comp_3# = app > map_1 > comp_2 = cons_2 > app# = comp_1 = cons_1 > map_2# = comp_0 = cons_0 partial status: pi(app#) = [] pi(comp_2) = [2] pi(comp_3#) = [2] pi(map_1) = [1] pi(map_2#) = [1] pi(cons_2) = [] pi(app) = [] pi(comp_0) = [] pi(comp_1) = [] pi(cons_0) = [] pi(cons_1) = [] The following rules are removed: p3 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y) p2: comp_3#(f,g,x) -> app#(g,x) p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p4: map_2#(f,cons_2(x,xs)) -> app#(f,x) p5: comp_3#(f,g,x) -> app#(f,app(g,x)) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The estimated dependency graph contains the following SCCs: {p3} {p1, p2, p5} -- 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(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The set of usable rules consists of r2, r4, r6 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: map_2#_A(x1,x2) = x2 + 11 cons_2_A(x1,x2) = max{x1 - 9, x2 + 12} app_A(x1,x2) = max{8, x1 + 2, x2 - 9} comp_1_A(x1) = x1 + 8 comp_2_A(x1,x2) = max{9, x1 + 4, x2 - 10} cons_0_A = 1 cons_1_A(x1) = 4 map_0_A = 1 map_1_A(x1) = max{4, x1 - 14} base order 2: precedence: cons_2 = app > map_2# = cons_1 = map_1 > comp_2 = cons_0 = map_0 > comp_1 partial status: pi(map_2#) = [2] pi(cons_2) = [2] pi(app) = [1] pi(comp_1) = [1] pi(comp_2) = [1] pi(cons_0) = [] pi(cons_1) = [] pi(map_0) = [] pi(map_1) = [] 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: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y) p2: comp_3#(f,g,x) -> app#(f,app(g,x)) p3: comp_3#(f,g,x) -> app#(g,x) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The set of usable rules consists of r1, r2, r4, r5, r6 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: app#_A(x1,x2) = max{2, x1 - 4} comp_2_A(x1,x2) = max{x1 + 9, x2} comp_3#_A(x1,x2,x3) = max{4, x1 + 1, x2 - 4} app_A(x1,x2) = max{25, x1 + 10, x2} comp_0_A = 1 comp_1_A(x1) = max{2, x1 - 1} cons_0_A = 1 cons_1_A(x1) = max{12, x1 - 27} cons_2_A(x1,x2) = max{23, x1 - 26, x2 - 51} map_0_A = 1 map_1_A(x1) = max{12, x1 - 29} base order 2: precedence: app = cons_0 > comp_2 = cons_2 > app# = comp_3# = comp_1 = cons_1 = map_1 > comp_0 = map_0 partial status: pi(app#) = [] pi(comp_2) = [] pi(comp_3#) = [] pi(app) = [] pi(comp_0) = [] pi(comp_1) = [] pi(cons_0) = [] pi(cons_1) = [] pi(cons_2) = [] pi(map_0) = [] pi(map_1) = [] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y) p2: comp_3#(f,g,x) -> app#(g,x) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(comp_2(x1,x2),y) -> comp_3#(x1,x2,y) p2: comp_3#(f,g,x) -> app#(g,x) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The set of usable rules consists of r2, r6 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: app#_A(x1,x2) = x1 comp_2_A(x1,x2) = x2 + 1 comp_3#_A(x1,x2,x3) = x2 + 1 app_A(x1,x2) = max{7, x1 - 10, x2 + 2} comp_1_A(x1) = max{8, x1 + 3} map_0_A = 5 map_1_A(x1) = max{6, x1 - 1} base order 2: precedence: comp_1 > app > app# = comp_2 = map_1 > comp_3# = map_0 partial status: pi(app#) = [1] pi(comp_2) = [2] pi(comp_3#) = [] pi(app) = [] pi(comp_1) = [] pi(map_0) = [] pi(map_1) = [] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: comp_3#(f,g,x) -> app#(g,x) and R consists of: r1: app(comp_0(),y) -> comp_1(y) r2: app(comp_1(x1),y) -> comp_2(x1,y) r3: app(comp_2(x1,x2),y) -> comp_3(x1,x2,y) r4: app(cons_0(),y) -> cons_1(y) r5: app(cons_1(x1),y) -> cons_2(x1,y) r6: app(map_0(),y) -> map_1(y) r7: app(map_1(x1),y) -> map_2(x1,y) r8: app(twice_0(),y) -> twice_1(y) r9: map_2(f,nil_0()) -> nil_0() r10: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r11: comp_3(f,g,x) -> app(f,app(g,x)) r12: twice_1(f) -> comp_2(f,f) The estimated dependency graph contains the following SCCs: (no SCCs)