YES We show the termination of the TRS R: minus(x,|0|()) -> x minus(s(x),s(y)) -> minus(x,y) quot(|0|(),s(y)) -> |0|() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) le(|0|(),y) -> true() le(s(x),|0|()) -> false() le(s(x),s(y)) -> le(x,y) app(nil(),y) -> y app(add(n,x),y) -> add(n,app(x,y)) low(n,nil()) -> nil() low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x)) if_low(true(),n,add(m,x)) -> add(m,low(n,x)) if_low(false(),n,add(m,x)) -> low(n,x) high(n,nil()) -> nil() high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x)) if_high(true(),n,add(m,x)) -> high(n,x) if_high(false(),n,add(m,x)) -> add(m,high(n,x)) quicksort(nil()) -> nil() quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x)))) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: minus#(s(x),s(y)) -> minus#(x,y) p2: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) p3: quot#(s(x),s(y)) -> minus#(x,y) p4: le#(s(x),s(y)) -> le#(x,y) p5: app#(add(n,x),y) -> app#(x,y) p6: low#(n,add(m,x)) -> if_low#(le(m,n),n,add(m,x)) p7: low#(n,add(m,x)) -> le#(m,n) p8: if_low#(true(),n,add(m,x)) -> low#(n,x) p9: if_low#(false(),n,add(m,x)) -> low#(n,x) p10: high#(n,add(m,x)) -> if_high#(le(m,n),n,add(m,x)) p11: high#(n,add(m,x)) -> le#(m,n) p12: if_high#(true(),n,add(m,x)) -> high#(n,x) p13: if_high#(false(),n,add(m,x)) -> high#(n,x) p14: quicksort#(add(n,x)) -> app#(quicksort(low(n,x)),add(n,quicksort(high(n,x)))) p15: quicksort#(add(n,x)) -> quicksort#(low(n,x)) p16: quicksort#(add(n,x)) -> low#(n,x) p17: quicksort#(add(n,x)) -> quicksort#(high(n,x)) p18: quicksort#(add(n,x)) -> high#(n,x) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: quot(|0|(),s(y)) -> |0|() r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r5: le(|0|(),y) -> true() r6: le(s(x),|0|()) -> false() r7: le(s(x),s(y)) -> le(x,y) r8: app(nil(),y) -> y r9: app(add(n,x),y) -> add(n,app(x,y)) r10: low(n,nil()) -> nil() r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x)) r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x)) r13: if_low(false(),n,add(m,x)) -> low(n,x) r14: high(n,nil()) -> nil() r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x)) r16: if_high(true(),n,add(m,x)) -> high(n,x) r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x)) r18: quicksort(nil()) -> nil() r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x)))) The estimated dependency graph contains the following SCCs: {p2} {p1} {p15, p17} {p10, p12, p13} {p6, p8, p9} {p4} {p5} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: quot(|0|(),s(y)) -> |0|() r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r5: le(|0|(),y) -> true() r6: le(s(x),|0|()) -> false() r7: le(s(x),s(y)) -> le(x,y) r8: app(nil(),y) -> y r9: app(add(n,x),y) -> add(n,app(x,y)) r10: low(n,nil()) -> nil() r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x)) r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x)) r13: if_low(false(),n,add(m,x)) -> low(n,x) r14: high(n,nil()) -> nil() r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x)) r16: if_high(true(),n,add(m,x)) -> high(n,x) r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x)) r18: quicksort(nil()) -> nil() r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x)))) The set of usable rules consists of r1, r2 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: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: quot#_A(x1,x2) = x1 s_A(x1) = x1 + 2 minus_A(x1,x2) = x1 + 1 |0|_A() = 2 quot#!_A(x1,x2) = 7 s!_A(x1) = 9 minus!_A(x1,x2) = 8 |0|!_A() = 16 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: quot#_A(x1,x2) = x1 s_A(x1) = x1 + 3 minus_A(x1,x2) = x1 + 1 |0|_A() = 5 quot#!_A(x1,x2) = x1 + 2 s!_A(x1) = x1 + 4 minus!_A(x1,x2) = x1 + 3 |0|!_A() = 3 partial status: pi(quot#) = [] pi(s) = [1] pi(minus) = [1] pi(|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: minus#(s(x),s(y)) -> minus#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: quot(|0|(),s(y)) -> |0|() r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r5: le(|0|(),y) -> true() r6: le(s(x),|0|()) -> false() r7: le(s(x),s(y)) -> le(x,y) r8: app(nil(),y) -> y r9: app(add(n,x),y) -> add(n,app(x,y)) r10: low(n,nil()) -> nil() r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x)) r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x)) r13: if_low(false(),n,add(m,x)) -> low(n,x) r14: high(n,nil()) -> nil() r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x)) r16: if_high(true(),n,add(m,x)) -> high(n,x) r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x)) r18: quicksort(nil()) -> nil() r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x)))) The set of usable rules consists of r10 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: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: minus#_A(x1,x2) = x1 + x2 s_A(x1) = x1 + 1 low_A(x1,x2) = x2 + 1 nil_A() = 3 minus#!_A(x1,x2) = 0 s!_A(x1) = 9 low!_A(x1,x2) = 1 nil!_A() = 0 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: minus#_A(x1,x2) = x1 + x2 s_A(x1) = x1 + 2 low_A(x1,x2) = x2 + 5 nil_A() = 7 minus#!_A(x1,x2) = x2 + 17 s!_A(x1) = 16 low!_A(x1,x2) = 9 nil!_A() = 6 partial status: pi(minus#) = [] pi(s) = [1] pi(low) = [] pi(nil) = [] 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: quicksort#(add(n,x)) -> quicksort#(high(n,x)) p2: quicksort#(add(n,x)) -> quicksort#(low(n,x)) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: quot(|0|(),s(y)) -> |0|() r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r5: le(|0|(),y) -> true() r6: le(s(x),|0|()) -> false() r7: le(s(x),s(y)) -> le(x,y) r8: app(nil(),y) -> y r9: app(add(n,x),y) -> add(n,app(x,y)) r10: low(n,nil()) -> nil() r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x)) r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x)) r13: if_low(false(),n,add(m,x)) -> low(n,x) r14: high(n,nil()) -> nil() r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x)) r16: if_high(true(),n,add(m,x)) -> high(n,x) r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x)) r18: quicksort(nil()) -> nil() r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x)))) The set of usable rules consists of r5, r6, r7, r10, r11, r12, r13, r14, r15, r16, r17 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: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: quicksort#_A(x1) = 0 add_A(x1,x2) = x1 + x2 high_A(x1,x2) = x2 low_A(x1,x2) = x2 + 1 le_A(x1,x2) = x2 + 1 |0|_A() = 2 true_A() = 1 s_A(x1) = x1 + 3 false_A() = 1 nil_A() = 2 if_low_A(x1,x2,x3) = x3 + 1 if_high_A(x1,x2,x3) = x3 quicksort#!_A(x1) = 3 add!_A(x1,x2) = 3 high!_A(x1,x2) = x1 + x2 low!_A(x1,x2) = 3 le!_A(x1,x2) = x1 + x2 |0|!_A() = 5 true!_A() = 2 s!_A(x1) = 7 false!_A() = 4 nil!_A() = 0 if_low!_A(x1,x2,x3) = 3 if_high!_A(x1,x2,x3) = x2 + x3 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: quicksort#_A(x1) = 0 add_A(x1,x2) = x1 + x2 + 6 high_A(x1,x2) = x2 low_A(x1,x2) = x2 le_A(x1,x2) = x1 + x2 + 1 |0|_A() = 2 true_A() = 1 s_A(x1) = x1 + 1 false_A() = 1 nil_A() = 2 if_low_A(x1,x2,x3) = x3 if_high_A(x1,x2,x3) = x3 quicksort#!_A(x1) = x1 + 3 add!_A(x1,x2) = x2 + 4 high!_A(x1,x2) = x1 + x2 + 3 low!_A(x1,x2) = x2 + 1 le!_A(x1,x2) = x2 + 8 |0|!_A() = 6 true!_A() = 6 s!_A(x1) = 7 false!_A() = 5 nil!_A() = 0 if_low!_A(x1,x2,x3) = x3 if_high!_A(x1,x2,x3) = x2 + x3 + 2 partial status: pi(quicksort#) = [] pi(add) = [] pi(high) = [1] pi(low) = [] pi(le) = [2] pi(|0|) = [] pi(true) = [] pi(s) = [1] pi(false) = [] pi(nil) = [] pi(if_low) = [] pi(if_high) = [1, 2] The following rules are removed: p1, p2 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: if_high#(false(),n,add(m,x)) -> high#(n,x) p2: high#(n,add(m,x)) -> if_high#(le(m,n),n,add(m,x)) p3: if_high#(true(),n,add(m,x)) -> high#(n,x) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: quot(|0|(),s(y)) -> |0|() r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r5: le(|0|(),y) -> true() r6: le(s(x),|0|()) -> false() r7: le(s(x),s(y)) -> le(x,y) r8: app(nil(),y) -> y r9: app(add(n,x),y) -> add(n,app(x,y)) r10: low(n,nil()) -> nil() r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x)) r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x)) r13: if_low(false(),n,add(m,x)) -> low(n,x) r14: high(n,nil()) -> nil() r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x)) r16: if_high(true(),n,add(m,x)) -> high(n,x) r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x)) r18: quicksort(nil()) -> nil() r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x)))) The set of usable rules consists of r5, r6, r7 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: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: if_high#_A(x1,x2,x3) = 0 false_A() = 3 add_A(x1,x2) = x1 high#_A(x1,x2) = 0 le_A(x1,x2) = x1 + 1 true_A() = 3 |0|_A() = 2 s_A(x1) = x1 + 2 if_high#!_A(x1,x2,x3) = x1 + 3 false!_A() = 4 add!_A(x1,x2) = x2 + 8 high#!_A(x1,x2) = x2 + 5 le!_A(x1,x2) = 5 true!_A() = 2 |0|!_A() = 3 s!_A(x1) = 6 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: if_high#_A(x1,x2,x3) = x3 false_A() = 3 add_A(x1,x2) = x1 + x2 high#_A(x1,x2) = x2 le_A(x1,x2) = x1 + 1 true_A() = 3 |0|_A() = 2 s_A(x1) = x1 + 2 if_high#!_A(x1,x2,x3) = x1 + 4 false!_A() = 5 add!_A(x1,x2) = x2 + 9 high#!_A(x1,x2) = x2 + 6 le!_A(x1,x2) = 0 true!_A() = 1 |0|!_A() = 2 s!_A(x1) = 4 partial status: pi(if_high#) = [1, 3] pi(false) = [] pi(add) = [2] pi(high#) = [2] pi(le) = [] pi(true) = [] pi(|0|) = [] pi(s) = [] The following rules are removed: p1, p2, p3 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: if_low#(false(),n,add(m,x)) -> low#(n,x) p2: low#(n,add(m,x)) -> if_low#(le(m,n),n,add(m,x)) p3: if_low#(true(),n,add(m,x)) -> low#(n,x) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: quot(|0|(),s(y)) -> |0|() r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r5: le(|0|(),y) -> true() r6: le(s(x),|0|()) -> false() r7: le(s(x),s(y)) -> le(x,y) r8: app(nil(),y) -> y r9: app(add(n,x),y) -> add(n,app(x,y)) r10: low(n,nil()) -> nil() r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x)) r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x)) r13: if_low(false(),n,add(m,x)) -> low(n,x) r14: high(n,nil()) -> nil() r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x)) r16: if_high(true(),n,add(m,x)) -> high(n,x) r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x)) r18: quicksort(nil()) -> nil() r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x)))) The set of usable rules consists of r6 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: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: if_low#_A(x1,x2,x3) = x3 false_A() = 1 add_A(x1,x2) = x2 + 2 low#_A(x1,x2) = x2 + 2 le_A(x1,x2) = 1 true_A() = 2 s_A(x1) = x1 + 1 |0|_A() = 2 if_low#!_A(x1,x2,x3) = 5 false!_A() = 2 add!_A(x1,x2) = 7 low#!_A(x1,x2) = 5 le!_A(x1,x2) = 4 true!_A() = 6 s!_A(x1) = 3 |0|!_A() = 7 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: if_low#_A(x1,x2,x3) = x3 false_A() = 1 add_A(x1,x2) = x1 + x2 + 2 low#_A(x1,x2) = x2 + 2 le_A(x1,x2) = x2 + 1 true_A() = 2 s_A(x1) = 2 |0|_A() = 2 if_low#!_A(x1,x2,x3) = x2 + x3 + 5 false!_A() = 0 add!_A(x1,x2) = 5 low#!_A(x1,x2) = x1 + x2 + 6 le!_A(x1,x2) = 2 true!_A() = 0 s!_A(x1) = 1 |0|!_A() = 2 partial status: pi(if_low#) = [3] pi(false) = [] pi(add) = [2] pi(low#) = [2] pi(le) = [] pi(true) = [] pi(s) = [1] pi(|0|) = [] The following rules are removed: p1, p2, p3 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: le#(s(x),s(y)) -> le#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: quot(|0|(),s(y)) -> |0|() r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r5: le(|0|(),y) -> true() r6: le(s(x),|0|()) -> false() r7: le(s(x),s(y)) -> le(x,y) r8: app(nil(),y) -> y r9: app(add(n,x),y) -> add(n,app(x,y)) r10: low(n,nil()) -> nil() r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x)) r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x)) r13: if_low(false(),n,add(m,x)) -> low(n,x) r14: high(n,nil()) -> nil() r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x)) r16: if_high(true(),n,add(m,x)) -> high(n,x) r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x)) r18: quicksort(nil()) -> nil() r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x)))) The set of usable rules consists of r3, r10 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: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: le#_A(x1,x2) = x1 + x2 s_A(x1) = x1 + 1 quot_A(x1,x2) = x2 |0|_A() = 1 low_A(x1,x2) = x2 + 1 nil_A() = 3 le#!_A(x1,x2) = 0 s!_A(x1) = 9 quot!_A(x1,x2) = 7 |0|!_A() = 6 low!_A(x1,x2) = 1 nil!_A() = 0 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: le#_A(x1,x2) = x1 + x2 s_A(x1) = x1 + 2 quot_A(x1,x2) = 4 |0|_A() = 4 low_A(x1,x2) = x2 + 5 nil_A() = 7 le#!_A(x1,x2) = x2 + 17 s!_A(x1) = 16 quot!_A(x1,x2) = 15 |0|!_A() = 14 low!_A(x1,x2) = 9 nil!_A() = 6 partial status: pi(le#) = [] pi(s) = [1] pi(quot) = [2] pi(|0|) = [] pi(low) = [] pi(nil) = [] 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#(add(n,x),y) -> app#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: quot(|0|(),s(y)) -> |0|() r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r5: le(|0|(),y) -> true() r6: le(s(x),|0|()) -> false() r7: le(s(x),s(y)) -> le(x,y) r8: app(nil(),y) -> y r9: app(add(n,x),y) -> add(n,app(x,y)) r10: low(n,nil()) -> nil() r11: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x)) r12: if_low(true(),n,add(m,x)) -> add(m,low(n,x)) r13: if_low(false(),n,add(m,x)) -> low(n,x) r14: high(n,nil()) -> nil() r15: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x)) r16: if_high(true(),n,add(m,x)) -> high(n,x) r17: if_high(false(),n,add(m,x)) -> add(m,high(n,x)) r18: quicksort(nil()) -> nil() r19: quicksort(add(n,x)) -> app(quicksort(low(n,x)),add(n,quicksort(high(n,x)))) The set of usable rules consists of (no rules) 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: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = x1 add_A(x1,x2) = x1 + x2 + 1 app#!_A(x1,x2) = x1 add!_A(x1,x2) = x1 + 5 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = 0 add_A(x1,x2) = x1 + 3 app#!_A(x1,x2) = 0 add!_A(x1,x2) = 5 partial status: pi(app#) = [2] pi(add) = [] The following rules are removed: p1 Then no dependency pair remains.