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) pred(s(x)) -> x minus(x,|0|()) -> x minus(x,s(y)) -> pred(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#(x,s(y)) -> pred#(minus(x,y)) p3: minus#(x,s(y)) -> minus#(x,y) p4: gcd#(s(x),s(y)) -> if_gcd#(le(y,x),s(x),s(y)) p5: gcd#(s(x),s(y)) -> le#(y,x) p6: if_gcd#(true(),s(x),s(y)) -> gcd#(minus(x,y),s(y)) p7: if_gcd#(true(),s(x),s(y)) -> minus#(x,y) p8: if_gcd#(false(),s(x),s(y)) -> gcd#(minus(y,x),s(x)) p9: 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: pred(s(x)) -> x r5: minus(x,|0|()) -> x r6: minus(x,s(y)) -> pred(minus(x,y)) r7: gcd(|0|(),y) -> y r8: gcd(s(x),|0|()) -> s(x) r9: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y)) r10: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y)) r11: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x)) The estimated dependency graph contains the following SCCs: {p4, p6, p8} {p1} {p3} -- 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: pred(s(x)) -> x r5: minus(x,|0|()) -> x r6: minus(x,s(y)) -> pred(minus(x,y)) r7: gcd(|0|(),y) -> y r8: gcd(s(x),|0|()) -> s(x) r9: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y)) r10: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y)) r11: 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, r6 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): 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 + 6 gcd#_A(x1,x2) = x1 + x2 + 5 minus_A(x1,x2) = x1 + 1 le_A(x1,x2) = x1 + x2 + 1 true_A() = 1 |0|_A() = 1 pred_A(x1) = x1 if_gcd#!_A(x1,x2,x3) = x2 + x3 false!_A() = 7 s!_A(x1) = 9 gcd#!_A(x1,x2) = x1 + x2 + 1 minus!_A(x1,x2) = 11 le!_A(x1,x2) = x1 + x2 + 14 true!_A() = 0 |0|!_A() = 8 pred!_A(x1) = 10 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: if_gcd#_A(x1,x2,x3) = max{0, x1 - 51, x2 - 64, x3 - 75} false_A = 2 s_A(x1) = x1 + 37 gcd#_A(x1,x2) = max{0, x1 - 39, x2 - 75} minus_A(x1,x2) = max{13, x1 + 1, x2 - 39} le_A(x1,x2) = max{50, x1 - 36, x2 - 1} true_A = 2 |0|_A = 0 pred_A(x1) = max{2, x1 - 14} if_gcd#!_A(x1,x2,x3) = max{5, x1 - 64, x2 - 29, x3 - 41} false!_A = 3 s!_A(x1) = max{7, x1 + 2} gcd#!_A(x1,x2) = max{4, x1 - 28, x2 - 40} minus!_A(x1,x2) = max{6, x2 - 5} le!_A(x1,x2) = max{8, x1 - 1} true!_A = 1 |0|!_A = 2 pred!_A(x1) = max{33, x1 - 14} partial status: pi(if_gcd#) = [2, 3] pi(false) = [] pi(s) = [1] pi(gcd#) = [1, 2] pi(minus) = [1] pi(le) = [1, 2] pi(true) = [] pi(|0|) = [] pi(pred) = [1] 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: pred(s(x)) -> x r5: minus(x,|0|()) -> x r6: minus(x,s(y)) -> pred(minus(x,y)) r7: gcd(|0|(),y) -> y r8: gcd(s(x),|0|()) -> s(x) r9: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y)) r10: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y)) r11: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: le#_A(x1,x2) = x1 + x2 s_A(x1) = x1 + 2 le#!_A(x1,x2) = 0 s!_A(x1) = 2 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: le#_A(x1,x2) = max{x1, x2} s_A(x1) = x1 + 16 le#!_A(x1,x2) = max{3, x1 - 12, x2 - 17} s!_A(x1) = 2 partial status: pi(le#) = [] pi(s) = [] 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#(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: pred(s(x)) -> x r5: minus(x,|0|()) -> x r6: minus(x,s(y)) -> pred(minus(x,y)) r7: gcd(|0|(),y) -> y r8: gcd(s(x),|0|()) -> s(x) r9: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y)) r10: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y)) r11: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: minus#_A(x1,x2) = x2 s_A(x1) = x1 + 2 minus#!_A(x1,x2) = x2 s!_A(x1) = 4 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: minus#_A(x1,x2) = 0 s_A(x1) = 3 minus#!_A(x1,x2) = 0 s!_A(x1) = 5 partial status: pi(minus#) = [] pi(s) = [1] The following rules are removed: p1 Then no dependency pair remains.