YES We show the termination of the TRS R: app(app(.(),|1|()),x) -> x app(app(.(),x),|1|()) -> x app(app(.(),app(i(),x)),x) -> |1|() app(app(.(),x),app(i(),x)) -> |1|() app(app(.(),app(i(),y)),app(app(.(),y),z)) -> z app(app(.(),y),app(app(.(),app(i(),y)),z)) -> z app(i(),|1|()) -> |1|() app(i(),app(i(),x)) -> x 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(filter(),f),nil()) -> nil() app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs) app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs)) app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs) Instead we prove termination of the uncurried system: app(._0(),y) -> ._1(y) app(._1(x1),y) -> ._2(x1,y) app(cons_0(),y) -> cons_1(y) app(cons_1(x1),y) -> cons_2(x1,y) app(filter_0(),y) -> filter_1(y) app(filter_1(x1),y) -> filter_2(x1,y) app(filter2_0(),y) -> filter2_1(y) app(filter2_1(x1),y) -> filter2_2(x1,y) app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) app(i_0(),y) -> i_1(y) app(map_0(),y) -> map_1(y) app(map_1(x1),y) -> map_2(x1,y) ._2(|1|_0(),x) -> x ._2(x,|1|_0()) -> x ._2(i_1(x),x) -> |1|_0() ._2(x,i_1(x)) -> |1|_0() ._2(i_1(y),._2(y,z)) -> z ._2(y,._2(i_1(y),z)) -> z i_1(|1|_0()) -> |1|_0() i_1(i_1(x)) -> x map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) filter_2(f,nil_0()) -> nil_0() filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(._1(x1),y) -> ._2#(x1,y) p2: app#(filter_1(x1),y) -> filter_2#(x1,y) p3: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p4: app#(i_0(),y) -> i_1#(y) p5: app#(map_1(x1),y) -> map_2#(x1,y) p6: map_2#(f,cons_2(x,xs)) -> app#(f,x) p7: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p9: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p10: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p11: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p2, p3, p5, p6, p7, p8, p9, p10, p11} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(filter_1(x1),y) -> filter_2#(x1,y) p2: filter_2#(f,cons_2(x,xs)) -> app#(f,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: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p7: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p9: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r1, r3, r4, r5, r7, r8, r12, r16, r17, r20, r24 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: app#_A(x1,x2) = max{7, x1 - 13} filter_1_A(x1) = x1 + 22 filter_2#_A(x1,x2) = max{8, x1 + 6} cons_2_A(x1,x2) = max{6, x1 - 24, x2 - 24} map_1_A(x1) = max{12, x1 + 8} map_2#_A(x1,x2) = max{7, x1 - 5} filter2_3_A(x1,x2,x3) = max{x1 - 19, x2 + 22, x3 - 1} filter2_4#_A(x1,x2,x3,x4) = max{8, x2 + 6} false_0_A = 1 app_A(x1,x2) = max{x1 - 23, x2 + 22} true_0_A = 1 ._0_A = 11 ._1_A(x1) = max{12, x1 - 1} cons_0_A = 22 cons_1_A(x1) = max{12, x1 - 1} filter_0_A = 8 filter2_0_A = 1 filter2_1_A(x1) = max{2, x1 - 1} filter2_2_A(x1,x2) = max{3, x1 - 25, x2 - 1} map_0_A = 11 ._2_A(x1,x2) = max{13, x1 - 25, x2 - 24} i_1_A(x1) = max{23, x1 - 1} |1|_0_A = 0 filter_2_A(x1,x2) = max{10, x1 - 22, x2 - 7} nil_0_A = 6 base order 2: precedence: app = cons_0 = i_1 > filter_2# = filter2_4# = ._2 = filter_2 > app# = filter_1 = map_2# = false_0 = ._1 = |1|_0 > cons_2 = ._0 = filter_0 > map_1 = filter2_3 = true_0 = cons_1 = nil_0 > filter2_2 = map_0 > filter2_1 > filter2_0 partial status: pi(app#) = [] pi(filter_1) = [1] pi(filter_2#) = [1] pi(cons_2) = [] pi(map_1) = [1] pi(map_2#) = [] pi(filter2_3) = [2] pi(filter2_4#) = [2] pi(false_0) = [] pi(app) = [2] pi(true_0) = [] pi(._0) = [] pi(._1) = [] pi(cons_0) = [] pi(cons_1) = [] pi(filter_0) = [] pi(filter2_0) = [] pi(filter2_1) = [] pi(filter2_2) = [] pi(map_0) = [] pi(._2) = [] pi(i_1) = [] pi(|1|_0) = [] pi(filter_2) = [] pi(nil_0) = [] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: filter_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) p4: map_2#(f,cons_2(x,xs)) -> app#(f,x) p5: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p6: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) p7: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p8: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p3: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p4: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p5: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) p6: app#(map_1(x1),y) -> map_2#(x1,y) p7: map_2#(f,cons_2(x,xs)) -> app#(f,x) p8: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r1, r4, r5, r7, r8, r12, r16, r17, r20, r24 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: filter_2#_A(x1,x2) = max{33, x2 + 31} cons_2_A(x1,x2) = max{15, x1 + 5, x2 + 12} app#_A(x1,x2) = max{45, x2 + 33} filter2_3_A(x1,x2,x3) = max{45, x1 - 8, x2 + 2, x3 - 8} filter2_4#_A(x1,x2,x3,x4) = max{44, x4 + 32} true_0_A = 8 app_A(x1,x2) = max{x1 - 19, x2 + 17} false_0_A = 1 map_1_A(x1) = max{15, x1 - 10} map_2#_A(x1,x2) = max{45, x2 + 29} ._0_A = 6 ._1_A(x1) = max{7, x1 - 1} cons_1_A(x1) = x1 + 25 filter_0_A = 18 filter_1_A(x1) = max{5, x1 - 1} filter2_0_A = 4 filter2_1_A(x1) = max{5, x1 - 1} filter2_2_A(x1,x2) = max{4, x1 - 21, x2 - 1} map_0_A = 14 ._2_A(x1,x2) = max{17, x1 - 21, x2 - 17} i_1_A(x1) = max{16, x1 - 1} |1|_0_A = 0 filter_2_A(x1,x2) = max{9, x1 - 27, x2 - 17} nil_0_A = 0 base order 2: precedence: filter2_4# > app# = map_2# > cons_2 > filter2_3 = map_1 > app = cons_1 = map_0 = filter_2 = nil_0 > true_0 = filter_1 = filter2_2 = ._2 = i_1 > filter_2# = ._1 = filter_0 = filter2_1 = |1|_0 > false_0 = ._0 = filter2_0 partial status: pi(filter_2#) = [] pi(cons_2) = [2] pi(app#) = [2] pi(filter2_3) = [2] pi(filter2_4#) = [4] pi(true_0) = [] pi(app) = [] pi(false_0) = [] pi(map_1) = [] pi(map_2#) = [2] pi(._0) = [] pi(._1) = [] pi(cons_1) = [] pi(filter_0) = [] pi(filter_1) = [] pi(filter2_0) = [] pi(filter2_1) = [] pi(filter2_2) = [] pi(map_0) = [] pi(._2) = [] pi(i_1) = [] pi(|1|_0) = [] pi(filter_2) = [] pi(nil_0) = [] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p2: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p3: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p4: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) p5: app#(map_1(x1),y) -> map_2#(x1,y) 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(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p5, p6, p7} {p2, p3, p4} -- 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(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r3, r5, r7, r8, r9, r12, r16, r17, r20, r24 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{8, x2 - 8} cons_2_A(x1,x2) = max{x1 + 17, x2 + 7} app#_A(x1,x2) = max{9, x2 - 1} map_1_A(x1) = max{2, x1 - 21} app_A(x1,x2) = max{3, x1 - 23, x2 - 20} cons_0_A = 1 cons_1_A(x1) = max{2, x1 - 21} filter_0_A = 28 filter_1_A(x1) = max{4, x1 - 21} filter2_0_A = 1 filter2_1_A(x1) = max{2, x1 - 21} filter2_2_A(x1,x2) = max{1, x1 - 45, x2 - 21} filter2_3_A(x1,x2,x3) = max{3, x1 - 68, x2 - 44, x3 - 21} map_0_A = 1 ._2_A(x1,x2) = max{3, x1 - 22, x2 - 20} i_1_A(x1) = max{2, x1 - 21} |1|_0_A = 0 filter_2_A(x1,x2) = max{5, x1 - 22, x2 - 19} nil_0_A = 0 base order 2: precedence: cons_2 > filter_0 = filter_2 = nil_0 > app > map_2# = map_1 = cons_1 = filter_1 = filter2_3 = ._2 = i_1 > app# = cons_0 = filter2_2 = map_0 = |1|_0 > filter2_1 > filter2_0 partial status: pi(map_2#) = [] pi(cons_2) = [2] pi(app#) = [] pi(map_1) = [] pi(app) = [] pi(cons_0) = [] pi(cons_1) = [] pi(filter_0) = [] pi(filter_1) = [] pi(filter2_0) = [] pi(filter2_1) = [] pi(filter2_2) = [] pi(filter2_3) = [] pi(map_0) = [] pi(._2) = [] pi(i_1) = [] pi(|1|_0) = [] pi(filter_2) = [] 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)) -> app#(f,x) p2: app#(map_1(x1),y) -> map_2#(x1,y) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p1, p2} -- 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) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r1, r3, r5, r7, r8, r9, r16, r17, r20, r24 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: map_2#_A(x1,x2) = x1 + 3 cons_2_A(x1,x2) = max{5, x1 - 27, x2 - 10} app#_A(x1,x2) = x1 + 2 map_1_A(x1) = x1 + 4 app_A(x1,x2) = max{20, x1 - 5, x2 - 11} ._0_A = 6 ._1_A(x1) = max{4, x1 - 12} cons_0_A = 4 cons_1_A(x1) = max{11, x1 - 22} filter_0_A = 3 filter_1_A(x1) = max{4, x1 - 12} filter2_0_A = 4 filter2_1_A(x1) = max{1, x1 - 12} filter2_2_A(x1,x2) = max{2, x1 - 18, x2 - 12} filter2_3_A(x1,x2,x3) = max{3, x1 - 24, x2 - 18, x3 - 35} ._2_A(x1,x2) = max{2, x1 - 18} i_1_A(x1) = max{2, x1 - 12} |1|_0_A = 1 filter_2_A(x1,x2) = max{5, x1 - 18, x2 - 12} nil_0_A = 0 base order 2: precedence: filter_2 > app > app# = map_1 = ._1 = filter_0 > map_2# = cons_2 = ._0 = filter_1 = ._2 = i_1 = nil_0 > cons_1 = filter2_3 = |1|_0 > cons_0 = filter2_0 = filter2_2 > filter2_1 partial status: pi(map_2#) = [] pi(cons_2) = [] pi(app#) = [1] pi(map_1) = [1] pi(app) = [] pi(._0) = [] pi(._1) = [] pi(cons_0) = [] pi(cons_1) = [] pi(filter_0) = [] pi(filter_1) = [] pi(filter2_0) = [] pi(filter2_1) = [] pi(filter2_2) = [] pi(filter2_3) = [] pi(._2) = [] pi(i_1) = [] pi(|1|_0) = [] pi(filter_2) = [] pi(nil_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) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: (no SCCs) -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p2: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p3: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r1, r3, r4, r5, r7, r9, r12, r14, r16, r17, r18, r20, r24 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: filter2_4#_A(x1,x2,x3,x4) = max{x3 + 2, x4 + 2} true_0_A = 33 filter_2#_A(x1,x2) = max{1, x2 - 2} cons_2_A(x1,x2) = max{38, x1 + 5, x2 + 5} app_A(x1,x2) = max{32, x1 + 8, x2 + 30} false_0_A = 31 ._0_A = 1 ._1_A(x1) = max{10, x1 - 9} cons_0_A = 26 cons_1_A(x1) = max{33, x1 - 1} filter_0_A = 13 filter_1_A(x1) = max{20, x1 + 1} filter2_0_A = 1 filter2_1_A(x1) = 10 filter2_2_A(x1,x2) = max{x1 + 1, x2 + 31} filter2_3_A(x1,x2,x3) = max{20, x1 + 2, x2, x3 - 9} map_0_A = 1 map_1_A(x1) = max{31, x1 - 9} ._2_A(x1,x2) = max{31, x1 - 32, x2 + 1} |1|_0_A = 2 i_1_A(x1) = max{31, x1 - 3} filter_2_A(x1,x2) = max{29, x1, x2 - 39} nil_0_A = 1 base order 2: precedence: ._0 = filter2_0 > app = ._1 = cons_0 = filter2_1 = filter2_2 = ._2 = i_1 > |1|_0 > cons_2 = filter2_3 > filter2_4# = cons_1 = map_1 = filter_2 = nil_0 > true_0 = filter_1 = map_0 > filter_2# = filter_0 > false_0 partial status: pi(filter2_4#) = [] pi(true_0) = [] pi(filter_2#) = [] pi(cons_2) = [2] pi(app) = [1, 2] pi(false_0) = [] pi(._0) = [] pi(._1) = [] pi(cons_0) = [] pi(cons_1) = [] pi(filter_0) = [] pi(filter_1) = [1] pi(filter2_0) = [] pi(filter2_1) = [] pi(filter2_2) = [] pi(filter2_3) = [1, 2] pi(map_0) = [] pi(map_1) = [] pi(._2) = [] pi(|1|_0) = [] pi(i_1) = [] pi(filter_2) = [1] pi(nil_0) = [] The following rules are removed: p3 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p2: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p2: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r1, r4, r5, r7, r9, r12, r16, r17, r20, r24 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: filter2_4#_A(x1,x2,x3,x4) = max{x3 + 8, x4 + 3} true_0_A = 1 filter_2#_A(x1,x2) = max{2, x2 - 1} cons_2_A(x1,x2) = max{17, x1 + 10, x2 + 5} app_A(x1,x2) = max{x1 - 11, x2 + 6} ._0_A = 10 ._1_A(x1) = max{6, x1 - 1} cons_1_A(x1) = x1 + 29 filter_0_A = 3 filter_1_A(x1) = max{4, x1 - 25} filter2_0_A = 10 filter2_1_A(x1) = 4 filter2_2_A(x1,x2) = max{x1 + 20, x2 + 7} filter2_3_A(x1,x2,x3) = max{8, x1 - 1} map_0_A = 1 map_1_A(x1) = max{0, x1 - 6} ._2_A(x1,x2) = max{6, x1 - 12} i_1_A(x1) = max{5, x1 - 2} |1|_0_A = 1 filter_2_A(x1,x2) = max{9, x1 - 24, x2 - 27} nil_0_A = 1 base order 2: precedence: cons_2 > filter_2 > true_0 > filter_2# = app = ._0 = filter2_0 = nil_0 > filter2_4# = ._1 = cons_1 = filter_1 = filter2_2 = filter2_3 = map_1 = i_1 > filter_0 = filter2_1 = map_0 = ._2 > |1|_0 partial status: pi(filter2_4#) = [] pi(true_0) = [] pi(filter_2#) = [] pi(cons_2) = [2] pi(app) = [] pi(._0) = [] pi(._1) = [] pi(cons_1) = [] pi(filter_0) = [] pi(filter_1) = [] pi(filter2_0) = [] pi(filter2_1) = [] pi(filter2_2) = [] pi(filter2_3) = [] pi(map_0) = [] pi(map_1) = [] pi(._2) = [] pi(i_1) = [] pi(|1|_0) = [] pi(filter_2) = [] pi(nil_0) = [] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(filter_0(),y) -> filter_1(y) r6: app(filter_1(x1),y) -> filter_2(x1,y) r7: app(filter2_0(),y) -> filter2_1(y) r8: app(filter2_1(x1),y) -> filter2_2(x1,y) r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r11: app(i_0(),y) -> i_1(y) r12: app(map_0(),y) -> map_1(y) r13: app(map_1(x1),y) -> map_2(x1,y) r14: ._2(|1|_0(),x) -> x r15: ._2(x,|1|_0()) -> x r16: ._2(i_1(x),x) -> |1|_0() r17: ._2(x,i_1(x)) -> |1|_0() r18: ._2(i_1(y),._2(y,z)) -> z r19: ._2(y,._2(i_1(y),z)) -> z r20: i_1(|1|_0()) -> |1|_0() r21: i_1(i_1(x)) -> x r22: map_2(f,nil_0()) -> nil_0() r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r24: filter_2(f,nil_0()) -> nil_0() r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: (no SCCs)