YES We show the termination of the relative TRS R/S: R: le(|0|(),y) -> true() le(s(x),|0|()) -> false() le(s(x),s(y)) -> le(x,y) minus(|0|(),y) -> |0|() minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) if_minus(true(),s(x),y) -> |0|() if_minus(false(),s(x),y) -> s(minus(x,y)) quot(|0|(),s(y)) -> |0|() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) log(s(|0|())) -> |0|() log(s(s(x))) -> s(log(s(quot(x,s(s(|0|())))))) 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: le#(s(x),s(y)) -> le#(x,y) p2: minus#(s(x),y) -> if_minus#(le(s(x),y),s(x),y) p3: minus#(s(x),y) -> le#(s(x),y) p4: if_minus#(false(),s(x),y) -> minus#(x,y) p5: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) p6: quot#(s(x),s(y)) -> minus#(x,y) p7: log#(s(s(x))) -> log#(s(quot(x,s(s(|0|()))))) p8: log#(s(s(x))) -> quot#(x,s(s(|0|()))) and R consists of: r1: le(|0|(),y) -> true() r2: le(s(x),|0|()) -> false() r3: le(s(x),s(y)) -> le(x,y) r4: minus(|0|(),y) -> |0|() r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) r6: if_minus(true(),s(x),y) -> |0|() r7: if_minus(false(),s(x),y) -> s(minus(x,y)) r8: quot(|0|(),s(y)) -> |0|() r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r10: log(s(|0|())) -> |0|() r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|())))))) r12: rand(x) -> x r13: rand(x) -> rand(s(x)) The estimated dependency graph contains the following SCCs: {p7} {p5} {p2, p4} {p1} -- Reduction pair. Consider the non-minimal dependency pair problem (P, R), where P consists of p1: log#(s(s(x))) -> log#(s(quot(x,s(s(|0|()))))) and R consists of: r1: le(|0|(),y) -> true() r2: le(s(x),|0|()) -> false() r3: le(s(x),s(y)) -> le(x,y) r4: minus(|0|(),y) -> |0|() r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) r6: if_minus(true(),s(x),y) -> |0|() r7: if_minus(false(),s(x),y) -> s(minus(x,y)) r8: quot(|0|(),s(y)) -> |0|() r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r10: log(s(|0|())) -> |0|() r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|())))))) r12: rand(x) -> x r13: 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 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: log#_A(x1) = ((1,0),(0,0)) x1 s_A(x1) = x1 quot_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (0,1) |0|_A() = (0,1) le_A(x1,x2) = ((0,0),(1,0)) x1 true_A() = (0,0) false_A() = (0,0) minus_A(x1,x2) = ((1,0),(0,0)) x1 + (0,1) if_minus_A(x1,x2,x3) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (0,1) log_A(x1) = ((1,0),(0,0)) x1 + (1,1) rand_A(x1) = x1 + (1,1) precedence: log# = s = quot = |0| = le = true = false = minus = if_minus = log = rand partial status: pi(log#) = [] pi(s) = [] pi(quot) = [] pi(|0|) = [] pi(le) = [] pi(true) = [] pi(false) = [] pi(minus) = [] pi(if_minus) = [] pi(log) = [] pi(rand) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: log#_A(x1) = ((0,1),(0,0)) x1 + (1,3) s_A(x1) = ((0,1),(0,1)) x1 + (8,6) quot_A(x1,x2) = ((0,1),(0,1)) x1 + (3,0) |0|_A() = (1,4) le_A(x1,x2) = (2,6) true_A() = (1,4) false_A() = (2,5) minus_A(x1,x2) = ((1,1),(0,1)) x1 if_minus_A(x1,x2,x3) = x2 log_A(x1) = ((1,0),(1,0)) x1 + (3,0) rand_A(x1) = (7,5) precedence: log > minus > if_minus > quot = |0| > log# = s = le = false = rand > true partial status: pi(log#) = [] pi(s) = [] pi(quot) = [] pi(|0|) = [] pi(le) = [] pi(true) = [] pi(false) = [] pi(minus) = [] pi(if_minus) = [] pi(log) = [] 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: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) and R consists of: r1: le(|0|(),y) -> true() r2: le(s(x),|0|()) -> false() r3: le(s(x),s(y)) -> le(x,y) r4: minus(|0|(),y) -> |0|() r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) r6: if_minus(true(),s(x),y) -> |0|() r7: if_minus(false(),s(x),y) -> s(minus(x,y)) r8: quot(|0|(),s(y)) -> |0|() r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r10: log(s(|0|())) -> |0|() r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|())))))) r12: rand(x) -> x r13: 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 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) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,3) s_A(x1) = ((1,0),(0,0)) x1 minus_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,0),(1,1)) x2 + (0,3) le_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (2,4) |0|_A() = (0,1) true_A() = (1,2) false_A() = (1,2) if_minus_A(x1,x2,x3) = ((1,0),(1,0)) x2 + (0,2) quot_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (0,1) log_A(x1) = ((1,0),(1,0)) x1 + (0,1) rand_A(x1) = x1 + (1,0) precedence: quot# = s = minus = le = |0| = true = false = if_minus = quot = log = rand partial status: pi(quot#) = [] pi(s) = [] pi(minus) = [] pi(le) = [] pi(|0|) = [] pi(true) = [] pi(false) = [] pi(if_minus) = [] pi(quot) = [] pi(log) = [] 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 + ((0,0),(1,0)) x2 + (1,5) s_A(x1) = ((1,0),(0,0)) x1 + (8,13) minus_A(x1,x2) = ((1,0),(1,1)) x1 + (0,20) le_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (4,1) |0|_A() = (2,14) true_A() = (3,1) false_A() = (7,12) if_minus_A(x1,x2,x3) = ((1,0),(1,1)) x2 quot_A(x1,x2) = ((1,0),(1,0)) x1 + (0,12) log_A(x1) = ((1,0),(0,0)) x1 + (1,14) rand_A(x1) = (0,0) precedence: true > log > s = le = quot > minus = if_minus > false > |0| > quot# = rand partial status: pi(quot#) = [] pi(s) = [] pi(minus) = [1] pi(le) = [] pi(|0|) = [] pi(true) = [] pi(false) = [] pi(if_minus) = [2] pi(quot) = [] pi(log) = [] 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: if_minus#(false(),s(x),y) -> minus#(x,y) p2: minus#(s(x),y) -> if_minus#(le(s(x),y),s(x),y) and R consists of: r1: le(|0|(),y) -> true() r2: le(s(x),|0|()) -> false() r3: le(s(x),s(y)) -> le(x,y) r4: minus(|0|(),y) -> |0|() r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) r6: if_minus(true(),s(x),y) -> |0|() r7: if_minus(false(),s(x),y) -> s(minus(x,y)) r8: quot(|0|(),s(y)) -> |0|() r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r10: log(s(|0|())) -> |0|() r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|())))))) r12: rand(x) -> x r13: 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 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: if_minus#_A(x1,x2,x3) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + ((1,1),(1,0)) x3 + (1,2) false_A() = (2,2) s_A(x1) = ((1,0),(0,0)) x1 minus#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,1),(1,0)) x2 + (3,2) le_A(x1,x2) = ((0,0),(1,0)) x1 + (2,2) |0|_A() = (0,1) true_A() = (1,2) minus_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (0,1) if_minus_A(x1,x2,x3) = ((1,1),(1,1)) x2 + (0,1) quot_A(x1,x2) = x1 log_A(x1) = ((1,0),(0,0)) x1 + (1,1) rand_A(x1) = ((1,0),(1,1)) x1 + (1,1) precedence: if_minus# = false = s = minus# = le = |0| = true = minus = if_minus = quot = log = rand partial status: pi(if_minus#) = [] pi(false) = [] pi(s) = [] pi(minus#) = [] pi(le) = [] pi(|0|) = [] pi(true) = [] pi(minus) = [] pi(if_minus) = [] pi(quot) = [] pi(log) = [] pi(rand) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: if_minus#_A(x1,x2,x3) = ((0,1),(0,0)) x1 + ((1,1),(0,0)) x2 + x3 + (0,2) false_A() = (4,2) s_A(x1) = ((0,1),(1,0)) x1 + (5,2) minus#_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (3,2) le_A(x1,x2) = (6,2) |0|_A() = (0,1) true_A() = (1,2) minus_A(x1,x2) = x1 if_minus_A(x1,x2,x3) = x2 quot_A(x1,x2) = x1 log_A(x1) = x1 + (1,1) rand_A(x1) = (0,0) precedence: quot > log > minus > true > if_minus > s = |0| > le > if_minus# = false = minus# = rand partial status: pi(if_minus#) = [] pi(false) = [] pi(s) = [] pi(minus#) = [2] pi(le) = [] pi(|0|) = [] pi(true) = [] pi(minus) = [] pi(if_minus) = [] pi(quot) = [] pi(log) = [] pi(rand) = [] The next rules are strictly ordered: p2 We remove them from the problem. -- SCC decomposition. Consider the non-minimal dependency pair problem (P, R), where P consists of p1: if_minus#(false(),s(x),y) -> minus#(x,y) and R consists of: r1: le(|0|(),y) -> true() r2: le(s(x),|0|()) -> false() r3: le(s(x),s(y)) -> le(x,y) r4: minus(|0|(),y) -> |0|() r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) r6: if_minus(true(),s(x),y) -> |0|() r7: if_minus(false(),s(x),y) -> s(minus(x,y)) r8: quot(|0|(),s(y)) -> |0|() r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r10: log(s(|0|())) -> |0|() r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|())))))) r12: rand(x) -> x r13: rand(x) -> rand(s(x)) The estimated dependency graph contains the following SCCs: (no SCCs) -- Reduction pair. Consider the non-minimal dependency pair problem (P, R), where P consists of p1: le#(s(x),s(y)) -> le#(x,y) and R consists of: r1: le(|0|(),y) -> true() r2: le(s(x),|0|()) -> false() r3: le(s(x),s(y)) -> le(x,y) r4: minus(|0|(),y) -> |0|() r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) r6: if_minus(true(),s(x),y) -> |0|() r7: if_minus(false(),s(x),y) -> s(minus(x,y)) r8: quot(|0|(),s(y)) -> |0|() r9: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) r10: log(s(|0|())) -> |0|() r11: log(s(s(x))) -> s(log(s(quot(x,s(s(|0|())))))) r12: rand(x) -> x r13: 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 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: le#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 s_A(x1) = ((1,0),(0,0)) x1 le_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (2,3) |0|_A() = (0,1) true_A() = (1,3) false_A() = (0,0) minus_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,1)) x2 + (0,2) if_minus_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((0,0),(1,0)) x3 + (0,1) quot_A(x1,x2) = x1 + (0,1) log_A(x1) = ((1,0),(0,0)) x1 + (0,1) rand_A(x1) = x1 + (1,1) precedence: le# = s = le = |0| = true = false = minus = if_minus = quot = log = rand partial status: pi(le#) = [] pi(s) = [] pi(le) = [] pi(|0|) = [] pi(true) = [] pi(false) = [] pi(minus) = [] pi(if_minus) = [] pi(quot) = [] pi(log) = [] pi(rand) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: le#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 s_A(x1) = ((1,0),(1,0)) x1 + (6,2) le_A(x1,x2) = ((1,0),(0,0)) x2 + (3,3) |0|_A() = (1,2) true_A() = (2,3) false_A() = (5,1) minus_A(x1,x2) = ((1,0),(1,1)) x1 + (0,9) if_minus_A(x1,x2,x3) = x2 quot_A(x1,x2) = ((1,0),(1,0)) x1 + (1,2) log_A(x1) = ((1,1),(1,1)) x1 + (3,0) rand_A(x1) = (1,1) precedence: log > quot > false > le = minus > if_minus = rand > s > |0| > le# = true partial status: pi(le#) = [] pi(s) = [] pi(le) = [] pi(|0|) = [] pi(true) = [] pi(false) = [] pi(minus) = [1] pi(if_minus) = [2] pi(quot) = [] pi(log) = [] pi(rand) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains.