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: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: if_gcd#_A(x1,x2,x3) = ((0,0),(1,0)) x1 + x2 + x3 + (0,1) false_A() = (1,7) s_A(x1) = ((1,0),(1,1)) x1 + (0,5) gcd#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (0,4) minus_A(x1,x2) = x1 + (0,2) le_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (2,15) true_A() = (1,1) |0|_A() = (2,6) gcd_A(x1,x2) = x1 + x2 + (3,6) if_gcd_A(x1,x2,x3) = x2 + x3 + (3,6) rand_A(x1) = ((1,0),(1,0)) x1 + (1,1) precedence: false = minus = le = true = |0| = gcd = if_gcd > if_gcd# = s = 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) = [] The next rules are strictly ordered: p1 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)) p2: 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 estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. 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)) p2: 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: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: gcd#_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,4) s_A(x1) = x1 + (0,4) if_gcd#_A(x1,x2,x3) = x2 + ((1,0),(0,0)) x3 + (2,3) le_A(x1,x2) = ((1,0),(1,1)) x1 + (3,11) true_A() = (1,5) minus_A(x1,x2) = x1 + (0,2) |0|_A() = (2,2) false_A() = (1,1) gcd_A(x1,x2) = x1 + x2 + (4,2) if_gcd_A(x1,x2,x3) = x2 + x3 + (4,1) rand_A(x1) = ((1,0),(1,0)) x1 + (1,5) precedence: |0| > false > gcd# = if_gcd# = minus = rand > le > s = true = gcd = if_gcd partial status: pi(gcd#) = [1] pi(s) = [] pi(if_gcd#) = [2] pi(le) = [] pi(true) = [] pi(minus) = [] pi(|0|) = [] pi(false) = [] pi(gcd) = [] pi(if_gcd) = [] 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: 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: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: le#_A(x1,x2) = x2 + (0,1) s_A(x1) = x1 + (0,3) le_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (3,4) |0|_A() = (4,8) true_A() = (1,7) false_A() = (3,20) minus_A(x1,x2) = x1 + (0,1) gcd_A(x1,x2) = x1 + x2 + (2,2) if_gcd_A(x1,x2,x3) = x2 + x3 + (2,1) rand_A(x1) = ((1,0),(1,0)) x1 + (1,4) precedence: s = minus = gcd = if_gcd = rand > le# = le = |0| = true = false partial status: pi(le#) = [2] 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. -- 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: weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: minus#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (0,1) s_A(x1) = ((1,0),(1,1)) x1 + (0,5) le_A(x1,x2) = ((1,0),(1,1)) x1 + (3,13) |0|_A() = (2,1) true_A() = (3,11) false_A() = (1,2) minus_A(x1,x2) = ((1,0),(1,1)) x1 + (0,3) gcd_A(x1,x2) = x1 + x2 + (2,2) if_gcd_A(x1,x2,x3) = x2 + x3 + (2,1) rand_A(x1) = ((1,0),(1,0)) x1 + (1,6) precedence: s = false = minus = gcd = if_gcd = rand > minus# = |0| > le = true partial status: pi(minus#) = [2] pi(s) = [] 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.