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(x,|0|()) -> x minus(s(x),s(y)) -> minus(x,y) gcd(|0|(),y) -> y gcd(s(x),|0|()) -> s(x) gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y)) if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y)) if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x)) 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),s(y)) -> minus#(x,y) p3: gcd#(s(x),s(y)) -> if_gcd#(le(y,x),s(x),s(y)) p4: gcd#(s(x),s(y)) -> le#(y,x) p5: if_gcd#(true(),s(x),s(y)) -> gcd#(minus(x,y),s(y)) p6: if_gcd#(true(),s(x),s(y)) -> minus#(x,y) p7: if_gcd#(false(),s(x),s(y)) -> gcd#(minus(y,x),s(x)) p8: if_gcd#(false(),s(x),s(y)) -> minus#(y,x) 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(x,|0|()) -> x r5: minus(s(x),s(y)) -> minus(x,y) r6: gcd(|0|(),y) -> y r7: gcd(s(x),|0|()) -> s(x) r8: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y)) r9: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y)) r10: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x)) r11: rand(x) -> x r12: rand(x) -> rand(s(x)) The estimated dependency graph contains the following SCCs: {p3, p5, p7} {p1} {p2} -- Reduction pair. Consider the non-minimal dependency pair problem (P, R), where P consists of p1: if_gcd#(false(),s(x),s(y)) -> gcd#(minus(y,x),s(x)) p2: gcd#(s(x),s(y)) -> if_gcd#(le(y,x),s(x),s(y)) p3: if_gcd#(true(),s(x),s(y)) -> gcd#(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(x,|0|()) -> x r5: minus(s(x),s(y)) -> minus(x,y) r6: gcd(|0|(),y) -> y r7: gcd(s(x),|0|()) -> s(x) r8: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y)) r9: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y)) r10: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x)) r11: rand(x) -> x r12: rand(x) -> rand(s(x)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: if_gcd#_A(x1,x2,x3) = ((1,0),(1,1)) x2 + ((1,0),(1,1)) x3 + (2,1) false_A() = (1,0) s_A(x1) = ((1,0),(1,1)) x1 gcd#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,1) minus_A(x1,x2) = x1 le_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (3,0) true_A() = (3,0) |0|_A() = (2,1) gcd_A(x1,x2) = x1 + x2 + (2,1) if_gcd_A(x1,x2,x3) = x2 + x3 + (2,1) rand_A(x1) = ((1,0),(0,0)) x1 + (1,1) precedence: if_gcd# = false = s = gcd# = minus = le = true = |0| = gcd = if_gcd = rand partial status: pi(if_gcd#) = [] pi(false) = [] pi(s) = [] pi(gcd#) = [] pi(minus) = [] pi(le) = [] pi(true) = [] pi(|0|) = [] pi(gcd) = [] pi(if_gcd) = [] pi(rand) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: if_gcd#_A(x1,x2,x3) = ((1,0),(1,0)) x2 + x3 + (1,1) false_A() = (3,7) s_A(x1) = ((1,0),(1,1)) x1 + (4,3) gcd#_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (2,2) minus_A(x1,x2) = ((1,0),(1,1)) x1 + (1,1) le_A(x1,x2) = x2 + (2,10) true_A() = (3,11) |0|_A() = (2,4) gcd_A(x1,x2) = x1 + x2 + (1,8) if_gcd_A(x1,x2,x3) = x2 + ((1,0),(1,1)) x3 + (0,3) rand_A(x1) = (1,4) precedence: s = |0| > if_gcd > if_gcd# = le = rand > gcd# > gcd > false = minus = true partial status: pi(if_gcd#) = [] pi(false) = [] pi(s) = [] pi(gcd#) = [2] pi(minus) = [1] pi(le) = [] pi(true) = [] pi(|0|) = [] pi(gcd) = [] pi(if_gcd) = [2] pi(rand) = [] The next rules are strictly ordered: p1, p3 We remove them from the problem. -- SCC decomposition. Consider the non-minimal dependency pair problem (P, R), where P consists of p1: gcd#(s(x),s(y)) -> if_gcd#(le(y,x),s(x),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(x,|0|()) -> x r5: minus(s(x),s(y)) -> minus(x,y) r6: gcd(|0|(),y) -> y r7: gcd(s(x),|0|()) -> s(x) r8: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y)) r9: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y)) r10: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x)) r11: rand(x) -> x r12: 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(x,|0|()) -> x r5: minus(s(x),s(y)) -> minus(x,y) r6: gcd(|0|(),y) -> y r7: gcd(s(x),|0|()) -> s(x) r8: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y)) r9: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y)) r10: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x)) r11: rand(x) -> x r12: rand(x) -> rand(s(x)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: max/plus interpretations on natural numbers: le#_A(x1,x2) = max{x1 + 1, x2 + 1} s_A(x1) = max{2, x1} le_A(x1,x2) = max{x1 + 1, x2 + 1} |0|_A = 5 true_A = 4 false_A = 1 minus_A(x1,x2) = max{x1, x2} gcd_A(x1,x2) = max{4, x1 + 2, x2 + 2} if_gcd_A(x1,x2,x3) = max{x1 + 1, x2 + 2, x3 + 2} rand_A(x1) = max{6, x1 + 3} precedence: le = true = minus > gcd = if_gcd > s > |0| = false > rand > le# partial status: pi(le#) = [1, 2] pi(s) = [1] pi(le) = [1, 2] pi(|0|) = [] pi(true) = [] pi(false) = [] pi(minus) = [1] pi(gcd) = [] pi(if_gcd) = [] pi(rand) = [] 2. weighted path order base order: max/plus interpretations on natural numbers: le#_A(x1,x2) = max{x1 - 2, x2} s_A(x1) = max{14, x1 + 3} le_A(x1,x2) = max{16, x1 - 4, x2 - 4} |0|_A = 1 true_A = 2 false_A = 2 minus_A(x1,x2) = max{28, x1 + 14} gcd_A(x1,x2) = 14 if_gcd_A(x1,x2,x3) = 14 rand_A(x1) = 1 precedence: le = true = false = minus = gcd = if_gcd = rand > le# = s = |0| partial status: pi(le#) = [2] pi(s) = [1] pi(le) = [] pi(|0|) = [] pi(true) = [] pi(false) = [] pi(minus) = [] pi(gcd) = [] pi(if_gcd) = [] 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: le(|0|(),y) -> true() r2: le(s(x),|0|()) -> false() r3: le(s(x),s(y)) -> le(x,y) r4: minus(x,|0|()) -> x r5: minus(s(x),s(y)) -> minus(x,y) r6: gcd(|0|(),y) -> y r7: gcd(s(x),|0|()) -> s(x) r8: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y)) r9: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y)) r10: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x)) r11: rand(x) -> x r12: rand(x) -> rand(s(x)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: max/plus interpretations on natural numbers: minus#_A(x1,x2) = x1 + 1 s_A(x1) = max{4, x1} le_A(x1,x2) = max{x1 + 2, x2 + 1} |0|_A = 8 true_A = 7 false_A = 3 minus_A(x1,x2) = max{9, x1} gcd_A(x1,x2) = max{12, x1 + 3, x2 + 3} if_gcd_A(x1,x2,x3) = max{12, x1 + 1, x2 + 3, x3 + 3} rand_A(x1) = max{10, x1 + 5} precedence: minus# > |0| > true = false = minus = gcd = if_gcd > s > le > rand partial status: pi(minus#) = [1] pi(s) = [1] pi(le) = [1, 2] pi(|0|) = [] pi(true) = [] pi(false) = [] pi(minus) = [1] pi(gcd) = [] pi(if_gcd) = [] pi(rand) = [] 2. weighted path order base order: max/plus interpretations on natural numbers: minus#_A(x1,x2) = 0 s_A(x1) = 15 le_A(x1,x2) = max{19, x1 - 4, x2 + 4} |0|_A = 2 true_A = 3 false_A = 5 minus_A(x1,x2) = max{26, x1 + 11} gcd_A(x1,x2) = 16 if_gcd_A(x1,x2,x3) = 16 rand_A(x1) = 14 precedence: minus = gcd = if_gcd > minus# = s = le = true = false > |0| = rand partial status: pi(minus#) = [] pi(s) = [] pi(le) = [2] pi(|0|) = [] pi(true) = [] pi(false) = [] pi(minus) = [] pi(gcd) = [] pi(if_gcd) = [] pi(rand) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains.