YES We show the termination of the TRS 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)) -- SCC decomposition. Consider the 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)) The estimated dependency graph contains the following SCCs: {p3, p5, p7} {p1} {p2} -- Reduction pair. Consider the 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)) The set of usable rules consists of r1, r2, r3, r4, r5 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: if_gcd#_A(x1,x2,x3) = x2 + x3 false_A() = 1 s_A(x1) = x1 + 4 gcd#_A(x1,x2) = x1 + x2 + 1 minus_A(x1,x2) = x1 + 2 le_A(x1,x2) = 3 true_A() = 1 |0|_A() = 2 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: if_gcd#_A(x1,x2,x3) = 1 false_A = 0 s_A(x1) = x1 + 10 gcd#_A(x1,x2) = 1 minus_A(x1,x2) = x1 + 40 le_A(x1,x2) = max{7, x1 - 12, x2 - 31} true_A = 0 |0|_A = 11 if_gcd#!_A(x1,x2,x3) = max{8, x1 + 1} false!_A = 11 s!_A(x1) = 0 gcd#!_A(x1,x2) = max{0, x1 - 40, x2 - 2} minus!_A(x1,x2) = max{0, x1 - 36, x2 - 36} le!_A(x1,x2) = max{9, x2 - 35} true!_A = 0 |0|!_A = 12 partial status: pi(if_gcd#) = [] pi(false) = [] pi(s) = [1] pi(gcd#) = [] pi(minus) = [] pi(le) = [] pi(true) = [] 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: 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)) The set of usable rules consists of r1 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: le#_A(x1,x2) = x1 + x2 + 4 s_A(x1) = x1 + 3 le_A(x1,x2) = x1 + x2 + 4 |0|_A() = 2 true_A() = 1 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: le#_A(x1,x2) = max{x1, x2} s_A(x1) = max{3, x1 + 1} le_A(x1,x2) = max{2, x1 - 4, x2 - 2} |0|_A = 1 true_A = 0 le#!_A(x1,x2) = max{4, x1 - 4, x2 - 4} s!_A(x1) = max{3, x1 - 25} le!_A(x1,x2) = max{4, x1 - 7, x2 - 18} |0|!_A = 2 true!_A = 1 partial status: pi(le#) = [1] pi(s) = [1] pi(le) = [1] pi(|0|) = [] pi(true) = [] 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: 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)) The set of usable rules consists of r1 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: minus#_A(x1,x2) = x1 + x2 + 4 s_A(x1) = x1 + 3 le_A(x1,x2) = x1 + x2 + 4 |0|_A() = 2 true_A() = 1 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: minus#_A(x1,x2) = max{x1, x2} s_A(x1) = max{3, x1 + 1} le_A(x1,x2) = max{2, x1 - 4, x2 - 2} |0|_A = 1 true_A = 0 minus#!_A(x1,x2) = max{4, x1 - 4, x2 - 4} s!_A(x1) = max{3, x1 - 23} le!_A(x1,x2) = max{4, x1 - 12, x2 - 16} |0|!_A = 2 true!_A = 1 partial status: pi(minus#) = [1] pi(s) = [1] pi(le) = [1] pi(|0|) = [] pi(true) = [] The following rules are removed: p1 Then no dependency pair remains.