YES We show the termination of the relative TRS R/S: 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))) app(nil(),y) -> y app(add(n,x),y) -> add(n,app(x,y)) reverse(nil()) -> nil() reverse(add(n,x)) -> app(reverse(x),add(n,nil())) shuffle(nil()) -> nil() shuffle(add(n,x)) -> add(n,shuffle(reverse(x))) concat(leaf(),y) -> y concat(cons(u,v),y) -> cons(u,concat(v,y)) less_leaves(x,leaf()) -> false() less_leaves(leaf(),cons(w,z)) -> true() less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z)) S: rand(x) -> x rand(x) -> rand(s(x)) -- SCC decomposition. Consider the non-minimal 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: app#(add(n,x),y) -> app#(x,y) p5: reverse#(add(n,x)) -> app#(reverse(x),add(n,nil())) p6: reverse#(add(n,x)) -> reverse#(x) p7: shuffle#(add(n,x)) -> shuffle#(reverse(x)) p8: shuffle#(add(n,x)) -> reverse#(x) p9: concat#(cons(u,v),y) -> concat#(v,y) p10: less_leaves#(cons(u,v),cons(w,z)) -> less_leaves#(concat(u,v),concat(w,z)) p11: less_leaves#(cons(u,v),cons(w,z)) -> concat#(u,v) p12: less_leaves#(cons(u,v),cons(w,z)) -> concat#(w,z) 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: app(nil(),y) -> y r6: app(add(n,x),y) -> add(n,app(x,y)) r7: reverse(nil()) -> nil() r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil())) r9: shuffle(nil()) -> nil() r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x))) r11: concat(leaf(),y) -> y r12: concat(cons(u,v),y) -> cons(u,concat(v,y)) r13: less_leaves(x,leaf()) -> false() r14: less_leaves(leaf(),cons(w,z)) -> true() r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z)) r16: rand(x) -> x r17: rand(x) -> rand(s(x)) The estimated dependency graph contains the following SCCs: {p2} {p1} {p7} {p6} {p4} {p10} {p9} -- Reduction pair. Consider the non-minimal 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: app(nil(),y) -> y r6: app(add(n,x),y) -> add(n,app(x,y)) r7: reverse(nil()) -> nil() r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil())) r9: shuffle(nil()) -> nil() r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x))) r11: concat(leaf(),y) -> y r12: concat(cons(u,v),y) -> cons(u,concat(v,y)) r13: less_leaves(x,leaf()) -> false() r14: less_leaves(leaf(),cons(w,z)) -> true() r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z)) r16: rand(x) -> x r17: rand(x) -> rand(s(x)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: quot#_A(x1,x2) = x1 s_A(x1) = x1 minus_A(x1,x2) = x1 |0|_A() = (0,0) quot_A(x1,x2) = x1 + (1,1) app_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(0,1)) x2 + (4,2) nil_A() = (1,4) add_A(x1,x2) = ((0,1),(0,0)) x2 + (2,1) reverse_A(x1) = x1 + (12,2) shuffle_A(x1) = (13,4) concat_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (1,1) leaf_A() = (1,1) cons_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (2,1) less_leaves_A(x1,x2) = (0,0) false_A() = (0,0) true_A() = (0,0) rand_A(x1) = x1 + (1,1) precedence: reverse > concat = leaf = cons > quot > s > shuffle > quot# > minus = |0| = app = nil = add = less_leaves = false = true = rand partial status: pi(quot#) = [1] pi(s) = [1] pi(minus) = [1] pi(|0|) = [] pi(quot) = [1] pi(app) = [2] pi(nil) = [] pi(add) = [] pi(reverse) = [1] pi(shuffle) = [] pi(concat) = [2] pi(leaf) = [] pi(cons) = [] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: quot#_A(x1,x2) = ((1,0),(1,0)) x1 + (1,0) s_A(x1) = x1 + (3,0) minus_A(x1,x2) = ((1,1),(0,1)) x1 + (1,4) |0|_A() = (3,0) quot_A(x1,x2) = ((1,0),(0,0)) x1 + (0,1) app_A(x1,x2) = x2 + (4,2) nil_A() = (4,5) add_A(x1,x2) = (3,1) reverse_A(x1) = ((0,0),(1,0)) x1 + (8,0) shuffle_A(x1) = (9,2) concat_A(x1,x2) = (1,3) leaf_A() = (0,1) cons_A(x1,x2) = (0,1) less_leaves_A(x1,x2) = (2,2) false_A() = (0,0) true_A() = (0,0) rand_A(x1) = (0,0) precedence: leaf > concat = cons > less_leaves = false > quot > nil > reverse > app > s > |0| > minus = add > shuffle > quot# = true = rand partial status: pi(quot#) = [] pi(s) = [] pi(minus) = [1] pi(|0|) = [] pi(quot) = [] pi(app) = [] pi(nil) = [] pi(add) = [] pi(reverse) = [] pi(shuffle) = [] pi(concat) = [] pi(leaf) = [] pi(cons) = [] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the non-minimal 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: app(nil(),y) -> y r6: app(add(n,x),y) -> add(n,app(x,y)) r7: reverse(nil()) -> nil() r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil())) r9: shuffle(nil()) -> nil() r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x))) r11: concat(leaf(),y) -> y r12: concat(cons(u,v),y) -> cons(u,concat(v,y)) r13: less_leaves(x,leaf()) -> false() r14: less_leaves(leaf(),cons(w,z)) -> true() r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z)) r16: rand(x) -> x r17: rand(x) -> rand(s(x)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: minus#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1) s_A(x1) = x1 minus_A(x1,x2) = x1 |0|_A() = (0,0) quot_A(x1,x2) = x1 + ((0,1),(1,0)) x2 + (1,1) app_A(x1,x2) = ((0,1),(0,0)) x1 + x2 + (2,0) nil_A() = (0,0) add_A(x1,x2) = ((0,0),(0,1)) x2 + (1,0) reverse_A(x1) = ((0,1),(0,1)) x1 + (4,0) shuffle_A(x1) = ((0,0),(0,1)) x1 + (5,1) concat_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1) leaf_A() = (2,1) cons_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (3,1) less_leaves_A(x1,x2) = x2 + (1,0) false_A() = (1,1) true_A() = (1,1) rand_A(x1) = x1 + (1,0) precedence: leaf = false > |0| > less_leaves > concat = cons = true > reverse = shuffle > minus# > quot > s > minus = app = nil = add = rand partial status: pi(minus#) = [1, 2] pi(s) = [1] pi(minus) = [1] pi(|0|) = [] pi(quot) = [1] pi(app) = [2] pi(nil) = [] pi(add) = [] pi(reverse) = [] pi(shuffle) = [] pi(concat) = [1] pi(leaf) = [] pi(cons) = [1, 2] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: minus#_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(1,0)) x2 + (1,3) s_A(x1) = ((1,1),(1,1)) x1 + (3,2) minus_A(x1,x2) = x1 + (1,3) |0|_A() = (1,0) quot_A(x1,x2) = ((0,0),(0,1)) x1 + (2,0) app_A(x1,x2) = x2 + (0,2) nil_A() = (0,4) add_A(x1,x2) = (0,1) reverse_A(x1) = (0,3) shuffle_A(x1) = (1,2) concat_A(x1,x2) = ((1,1),(0,0)) x1 + (2,2) leaf_A() = (0,0) cons_A(x1,x2) = ((0,0),(1,0)) x2 + (1,1) less_leaves_A(x1,x2) = (0,0) false_A() = (0,0) true_A() = (0,0) rand_A(x1) = (1,1) precedence: concat = leaf = cons = less_leaves = false > reverse > app = add = shuffle > minus# = s > minus > quot > |0| = nil > true = rand partial status: pi(minus#) = [] pi(s) = [1] pi(minus) = [1] pi(|0|) = [] pi(quot) = [] pi(app) = [2] pi(nil) = [] pi(add) = [] pi(reverse) = [] pi(shuffle) = [] pi(concat) = [] pi(leaf) = [] pi(cons) = [] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the non-minimal dependency pair problem (P, R), where P consists of p1: shuffle#(add(n,x)) -> shuffle#(reverse(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: app(nil(),y) -> y r6: app(add(n,x),y) -> add(n,app(x,y)) r7: reverse(nil()) -> nil() r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil())) r9: shuffle(nil()) -> nil() r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x))) r11: concat(leaf(),y) -> y r12: concat(cons(u,v),y) -> cons(u,concat(v,y)) r13: less_leaves(x,leaf()) -> false() r14: less_leaves(leaf(),cons(w,z)) -> true() r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z)) r16: rand(x) -> x r17: rand(x) -> rand(s(x)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: shuffle#_A(x1) = ((1,1),(1,1)) x1 + (1,2) add_A(x1,x2) = ((0,0),(0,1)) x1 + x2 + (0,4) reverse_A(x1) = x1 + (3,0) minus_A(x1,x2) = x1 |0|_A() = (0,0) s_A(x1) = x1 quot_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,1),(1,0)) x2 + (1,1) app_A(x1,x2) = ((0,0),(0,1)) x1 + x2 + (1,0) nil_A() = (1,0) shuffle_A(x1) = ((1,1),(0,1)) x1 + (1,5) concat_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(0,1)) x2 + (2,1) leaf_A() = (1,1) cons_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (1,0) less_leaves_A(x1,x2) = (0,0) false_A() = (0,0) true_A() = (0,0) rand_A(x1) = x1 + (1,0) precedence: quot = shuffle > |0| = s > minus > shuffle# = reverse > app > add > nil > concat > leaf = cons = less_leaves = false = true = rand partial status: pi(shuffle#) = [1] pi(add) = [] pi(reverse) = [] pi(minus) = [1] pi(|0|) = [] pi(s) = [1] pi(quot) = [1] pi(app) = [] pi(nil) = [] pi(shuffle) = [1] pi(concat) = [2] pi(leaf) = [] pi(cons) = [2] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: shuffle#_A(x1) = x1 + (1,0) add_A(x1,x2) = (3,5) reverse_A(x1) = (1,4) minus_A(x1,x2) = x1 + (2,1) |0|_A() = (0,1) s_A(x1) = ((0,1),(0,0)) x1 + (1,0) quot_A(x1,x2) = ((0,1),(0,0)) x1 + (3,1) app_A(x1,x2) = (1,1) nil_A() = (3,4) shuffle_A(x1) = ((0,0),(1,0)) x1 concat_A(x1,x2) = ((1,1),(1,1)) x2 + (2,0) leaf_A() = (0,1) cons_A(x1,x2) = ((1,0),(1,1)) x2 + (0,1) less_leaves_A(x1,x2) = (1,2) false_A() = (0,1) true_A() = (0,0) rand_A(x1) = (2,1) precedence: nil = shuffle = concat > leaf > shuffle# = reverse = app > quot > |0| > minus > add > false > s > less_leaves > rand > cons = true partial status: pi(shuffle#) = [1] pi(add) = [] pi(reverse) = [] pi(minus) = [1] pi(|0|) = [] pi(s) = [] pi(quot) = [] pi(app) = [] pi(nil) = [] pi(shuffle) = [] pi(concat) = [] pi(leaf) = [] pi(cons) = [] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the non-minimal dependency pair problem (P, R), where P consists of p1: reverse#(add(n,x)) -> reverse#(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: app(nil(),y) -> y r6: app(add(n,x),y) -> add(n,app(x,y)) r7: reverse(nil()) -> nil() r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil())) r9: shuffle(nil()) -> nil() r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x))) r11: concat(leaf(),y) -> y r12: concat(cons(u,v),y) -> cons(u,concat(v,y)) r13: less_leaves(x,leaf()) -> false() r14: less_leaves(leaf(),cons(w,z)) -> true() r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z)) r16: rand(x) -> x r17: rand(x) -> rand(s(x)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: reverse#_A(x1) = ((1,1),(0,1)) x1 + (1,1) add_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (4,3) minus_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(1,1)) x2 + (2,0) |0|_A() = (0,0) s_A(x1) = ((0,1),(1,0)) x1 quot_A(x1,x2) = (1,1) app_A(x1,x2) = x1 + x2 nil_A() = (0,0) reverse_A(x1) = x1 + (1,1) shuffle_A(x1) = ((1,1),(1,0)) x1 + (1,3) concat_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (3,7) leaf_A() = (0,1) cons_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (4,2) less_leaves_A(x1,x2) = ((0,0),(1,0)) x2 + (2,2) false_A() = (1,2) true_A() = (1,2) rand_A(x1) = ((1,1),(1,1)) x1 + (1,0) precedence: shuffle = less_leaves > nil = reverse = true > minus = |0| = concat = cons > reverse# = add = s = quot = app = leaf = false = rand partial status: pi(reverse#) = [1] pi(add) = [] pi(minus) = [] pi(|0|) = [] pi(s) = [] pi(quot) = [] pi(app) = [2] pi(nil) = [] pi(reverse) = [] pi(shuffle) = [] pi(concat) = [2] pi(leaf) = [] pi(cons) = [] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: reverse#_A(x1) = (1,1) add_A(x1,x2) = (2,1) minus_A(x1,x2) = (0,1) |0|_A() = (0,0) s_A(x1) = (1,0) quot_A(x1,x2) = (1,0) app_A(x1,x2) = (2,1) nil_A() = (1,0) reverse_A(x1) = (4,3) shuffle_A(x1) = (3,2) concat_A(x1,x2) = ((1,1),(1,1)) x2 + (3,2) leaf_A() = (0,0) cons_A(x1,x2) = (0,0) less_leaves_A(x1,x2) = (2,1) false_A() = (1,1) true_A() = (0,0) rand_A(x1) = (0,0) precedence: concat > minus = quot = cons > leaf > |0| = s = shuffle = rand > reverse > nil > app > add = less_leaves > reverse# > false = true partial status: pi(reverse#) = [] pi(add) = [] pi(minus) = [] pi(|0|) = [] pi(s) = [] pi(quot) = [] pi(app) = [] pi(nil) = [] pi(reverse) = [] pi(shuffle) = [] pi(concat) = [2] pi(leaf) = [] pi(cons) = [] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the non-minimal 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: app(nil(),y) -> y r6: app(add(n,x),y) -> add(n,app(x,y)) r7: reverse(nil()) -> nil() r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil())) r9: shuffle(nil()) -> nil() r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x))) r11: concat(leaf(),y) -> y r12: concat(cons(u,v),y) -> cons(u,concat(v,y)) r13: less_leaves(x,leaf()) -> false() r14: less_leaves(leaf(),cons(w,z)) -> true() r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z)) r16: rand(x) -> x r17: rand(x) -> rand(s(x)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: app#_A(x1,x2) = ((1,1),(0,1)) x1 + x2 + (0,1) add_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (0,2) minus_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(1,1)) x2 + (2,2) |0|_A() = (0,0) s_A(x1) = ((0,1),(1,0)) x1 quot_A(x1,x2) = (1,1) app_A(x1,x2) = x1 + x2 nil_A() = (0,0) reverse_A(x1) = x1 + (1,0) shuffle_A(x1) = ((1,1),(0,1)) x1 + (1,3) concat_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1) leaf_A() = (2,1) cons_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (2,2) less_leaves_A(x1,x2) = ((0,0),(1,1)) x1 + (3,0) false_A() = (0,0) true_A() = (1,1) rand_A(x1) = ((1,1),(1,1)) x1 + (1,0) precedence: app# > minus = quot > |0| = s > shuffle > reverse > app > add > concat = leaf = cons = less_leaves = false > nil = true = rand partial status: pi(app#) = [1, 2] pi(add) = [2] pi(minus) = [] pi(|0|) = [] pi(s) = [] pi(quot) = [] pi(app) = [1, 2] pi(nil) = [] pi(reverse) = [1] pi(shuffle) = [1] pi(concat) = [] pi(leaf) = [] pi(cons) = [1] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: app#_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (1,0) add_A(x1,x2) = ((1,0),(0,0)) x2 + (2,5) minus_A(x1,x2) = (0,0) |0|_A() = (0,0) s_A(x1) = (0,0) quot_A(x1,x2) = (0,1) app_A(x1,x2) = ((1,0),(0,0)) x2 + (2,1) nil_A() = (3,1) reverse_A(x1) = ((1,1),(0,0)) x1 + (1,1) shuffle_A(x1) = ((0,1),(0,0)) x1 + (3,6) concat_A(x1,x2) = (3,2) leaf_A() = (0,0) cons_A(x1,x2) = ((1,0),(1,1)) x1 + (1,0) less_leaves_A(x1,x2) = (2,1) false_A() = (0,0) true_A() = (0,0) rand_A(x1) = (1,1) precedence: minus = |0| = s = quot = shuffle = concat = false = rand > reverse > add = app > nil = leaf > less_leaves = true > cons > app# partial status: pi(app#) = [] pi(add) = [] pi(minus) = [] pi(|0|) = [] pi(s) = [] pi(quot) = [] pi(app) = [] pi(nil) = [] pi(reverse) = [] pi(shuffle) = [] pi(concat) = [] pi(leaf) = [] pi(cons) = [1] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the non-minimal dependency pair problem (P, R), where P consists of p1: less_leaves#(cons(u,v),cons(w,z)) -> less_leaves#(concat(u,v),concat(w,z)) 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: app(nil(),y) -> y r6: app(add(n,x),y) -> add(n,app(x,y)) r7: reverse(nil()) -> nil() r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil())) r9: shuffle(nil()) -> nil() r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x))) r11: concat(leaf(),y) -> y r12: concat(cons(u,v),y) -> cons(u,concat(v,y)) r13: less_leaves(x,leaf()) -> false() r14: less_leaves(leaf(),cons(w,z)) -> true() r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z)) r16: rand(x) -> x r17: rand(x) -> rand(s(x)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: max/plus interpretations on natural numbers: less_leaves#_A(x1,x2) = max{8, x1 + 2, x2 + 1} cons_A(x1,x2) = max{x1 + 6, x2} concat_A(x1,x2) = max{x1 + 5, x2} minus_A(x1,x2) = x1 + 5 |0|_A = 0 s_A(x1) = x1 quot_A(x1,x2) = x2 + 3 app_A(x1,x2) = max{20, x1 - 9, x2} nil_A = 0 add_A(x1,x2) = max{9, x2 - 2} reverse_A(x1) = x1 + 12 shuffle_A(x1) = 10 leaf_A = 4 less_leaves_A(x1,x2) = max{x1 + 6, x2} false_A = 1 true_A = 3 rand_A(x1) = x1 + 2 precedence: cons = concat > |0| > minus = s = quot = app = shuffle = leaf > true > add = reverse > less_leaves# > nil = less_leaves = false > rand partial status: pi(less_leaves#) = [1, 2] pi(cons) = [1, 2] pi(concat) = [1, 2] pi(minus) = [] pi(|0|) = [] pi(s) = [] pi(quot) = [2] pi(app) = [2] pi(nil) = [] pi(add) = [] pi(reverse) = [] pi(shuffle) = [] pi(leaf) = [] pi(less_leaves) = [2] pi(false) = [] pi(true) = [] pi(rand) = [] 2. weighted path order base order: max/plus interpretations on natural numbers: less_leaves#_A(x1,x2) = max{x1 + 1, x2 + 1} cons_A(x1,x2) = max{x1 + 2, x2} concat_A(x1,x2) = max{x1, x2} minus_A(x1,x2) = 18 |0|_A = 14 s_A(x1) = 8 quot_A(x1,x2) = x2 + 10 app_A(x1,x2) = max{12, x2} nil_A = 16 add_A(x1,x2) = 1 reverse_A(x1) = 11 shuffle_A(x1) = 4 leaf_A = 0 less_leaves_A(x1,x2) = max{0, x2 - 3} false_A = 1 true_A = 1 rand_A(x1) = 7 precedence: less_leaves# = cons = concat = true > minus = |0| = s = quot = app = nil = add = reverse = shuffle = leaf = less_leaves = false = rand partial status: pi(less_leaves#) = [2] pi(cons) = [1, 2] pi(concat) = [1] pi(minus) = [] pi(|0|) = [] pi(s) = [] pi(quot) = [2] pi(app) = [2] pi(nil) = [] pi(add) = [] pi(reverse) = [] pi(shuffle) = [] pi(leaf) = [] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the non-minimal dependency pair problem (P, R), where P consists of p1: concat#(cons(u,v),y) -> concat#(v,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: app(nil(),y) -> y r6: app(add(n,x),y) -> add(n,app(x,y)) r7: reverse(nil()) -> nil() r8: reverse(add(n,x)) -> app(reverse(x),add(n,nil())) r9: shuffle(nil()) -> nil() r10: shuffle(add(n,x)) -> add(n,shuffle(reverse(x))) r11: concat(leaf(),y) -> y r12: concat(cons(u,v),y) -> cons(u,concat(v,y)) r13: less_leaves(x,leaf()) -> false() r14: less_leaves(leaf(),cons(w,z)) -> true() r15: less_leaves(cons(u,v),cons(w,z)) -> less_leaves(concat(u,v),concat(w,z)) r16: rand(x) -> x r17: rand(x) -> rand(s(x)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: max/plus interpretations on natural numbers: concat#_A(x1,x2) = max{x1 + 1, x2 + 1} cons_A(x1,x2) = max{x1 + 7, x2} minus_A(x1,x2) = max{x1 + 3, x2 - 1} |0|_A = 0 s_A(x1) = x1 quot_A(x1,x2) = x2 + 4 app_A(x1,x2) = max{5, x1, x2 + 2} nil_A = 8 add_A(x1,x2) = 1 reverse_A(x1) = 11 shuffle_A(x1) = 11 concat_A(x1,x2) = max{7, x1 + 4, x2} leaf_A = 6 less_leaves_A(x1,x2) = max{x1 - 4, x2 + 8} false_A = 1 true_A = 1 rand_A(x1) = x1 + 2 precedence: concat# > false = true > cons = minus = |0| = quot = app = nil = add = reverse = shuffle = concat = leaf = less_leaves > s = rand partial status: pi(concat#) = [1] pi(cons) = [1, 2] pi(minus) = [] pi(|0|) = [] pi(s) = [] pi(quot) = [2] pi(app) = [] pi(nil) = [] pi(add) = [] pi(reverse) = [] pi(shuffle) = [] pi(concat) = [1, 2] pi(leaf) = [] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] 2. weighted path order base order: max/plus interpretations on natural numbers: concat#_A(x1,x2) = max{6, x1 + 1} cons_A(x1,x2) = max{5, x1 + 3, x2} minus_A(x1,x2) = 10 |0|_A = 10 s_A(x1) = 4 quot_A(x1,x2) = max{9, x2 + 1} app_A(x1,x2) = 33 nil_A = 44 add_A(x1,x2) = 11 reverse_A(x1) = 33 shuffle_A(x1) = 16 concat_A(x1,x2) = max{x1 + 7, x2 + 7} leaf_A = 0 less_leaves_A(x1,x2) = 0 false_A = 1 true_A = 2 rand_A(x1) = 0 precedence: concat# > false > |0| = quot = nil = reverse = less_leaves > cons = minus = s = app = add = shuffle = concat = leaf = true > rand partial status: pi(concat#) = [1] pi(cons) = [1, 2] pi(minus) = [] pi(|0|) = [] pi(s) = [] pi(quot) = [2] pi(app) = [] pi(nil) = [] pi(add) = [] pi(reverse) = [] pi(shuffle) = [] pi(concat) = [1, 2] pi(leaf) = [] pi(less_leaves) = [] pi(false) = [] pi(true) = [] pi(rand) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains.