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!): max/plus interpretation on natural numbers: if_gcd#_A(x1,x2,x3) = max{0, x1 - 20, x2 - 32, x3 - 33} false_A = 7 s_A(x1) = max{17, x1 + 12} gcd#_A(x1,x2) = max{0, x1 - 32, x2 - 33} minus_A(x1,x2) = max{31, x1 + 9} le_A(x1,x2) = max{6, x1 - 11, x2 - 3} true_A = 2 |0|_A = 10 pred_A(x1) = max{10, x1 - 12} if_gcd#!_A(x1,x2,x3) = max{3, x1 - 15, x2 - 30, x3 - 31} false!_A = 19 s!_A(x1) = max{18, x1} gcd#!_A(x1,x2) = max{3, x1 - 29, x2 - 31} minus!_A(x1,x2) = max{27, x1 + 22, x2 - 18} le!_A(x1,x2) = 20 true!_A = 1 |0|!_A = 20 pred!_A(x1) = 23 2. 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() = 3 s_A(x1) = x1 + 7 gcd#_A(x1,x2) = x1 + x2 + 6 minus_A(x1,x2) = x1 + 1 le_A(x1,x2) = 5 true_A() = 1 |0|_A() = 0 pred_A(x1) = x1 if_gcd#!_A(x1,x2,x3) = x1 + x2 + x3 false!_A() = 19 s!_A(x1) = x1 + 18 gcd#!_A(x1,x2) = x1 + x2 + 6 minus!_A(x1,x2) = 16 le!_A(x1,x2) = x1 + 17 true!_A() = 21 |0|!_A() = 20 pred!_A(x1) = 15 partial status: pi(if_gcd#) = [] pi(false) = [] pi(s) = [1] pi(gcd#) = [] pi(minus) = [1, 2] pi(le) = [] 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!): max/plus interpretation on natural numbers: le#_A(x1,x2) = 0 s_A(x1) = max{7, x1 - 7} le#!_A(x1,x2) = 1 s!_A(x1) = 4 2. 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 + 5 le#!_A(x1,x2) = x1 + x2 s!_A(x1) = 4 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!): max/plus interpretation on natural numbers: minus#_A(x1,x2) = 0 s_A(x1) = max{5, x1 - 2} minus#!_A(x1,x2) = max{18, x1 + 1} s!_A(x1) = x1 + 17 2. 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 + 1 minus#!_A(x1,x2) = x2 s!_A(x1) = x1 + 4 partial status: pi(minus#) = [1] pi(s) = [] The following rules are removed: p1 Then no dependency pair remains.