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) mod(|0|(),y) -> |0|() mod(s(x),|0|()) -> |0|() mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) if_mod(false(),s(x),s(y)) -> 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: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y)) p4: mod#(s(x),s(y)) -> le#(y,x) p5: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y)) p6: if_mod#(true(),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: mod(|0|(),y) -> |0|() r7: mod(s(x),|0|()) -> |0|() r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) r10: if_mod(false(),s(x),s(y)) -> s(x) The estimated dependency graph contains the following SCCs: {p3, p5} {p1} {p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y)) p2: mod#(s(x),s(y)) -> if_mod#(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: mod(|0|(),y) -> |0|() r7: mod(s(x),|0|()) -> |0|() r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) r10: if_mod(false(),s(x),s(y)) -> 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: 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_mod#_A(x1,x2,x3) = x3 true_A() = 1 s_A(x1) = x1 + 1 mod#_A(x1,x2) = x2 minus_A(x1,x2) = x1 + 1 le_A(x1,x2) = x2 + 1 |0|_A() = 2 false_A() = 1 if_mod#!_A(x1,x2,x3) = 1 true!_A() = 0 s!_A(x1) = 3 mod#!_A(x1,x2) = 2 minus!_A(x1,x2) = 2 le!_A(x1,x2) = 1 |0|!_A() = 1 false!_A() = 0 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: if_mod#_A(x1,x2,x3) = max{0, x1 - 43, x2 - 70, x3 - 28} true_A = 0 s_A(x1) = x1 + 27 mod#_A(x1,x2) = max{0, x1 - 52, x2 - 28} minus_A(x1,x2) = max{10, x1 + 9} le_A(x1,x2) = max{21, x1 - 28, x2 - 28} |0|_A = 10 false_A = 21 if_mod#!_A(x1,x2,x3) = max{1, x2 - 1} true!_A = 8 s!_A(x1) = x1 + 19 mod#!_A(x1,x2) = max{18, x1 + 6} minus!_A(x1,x2) = max{2, x1 - 1} le!_A(x1,x2) = max{9, x1 - 3} |0|!_A = 7 false!_A = 6 partial status: pi(if_mod#) = [2] pi(true) = [] pi(s) = [1] pi(mod#) = [1] pi(minus) = [1] pi(le) = [] pi(|0|) = [] pi(false) = [] The following rules are removed: p1, p2 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: mod(|0|(),y) -> |0|() r7: mod(s(x),|0|()) -> |0|() r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) r10: if_mod(false(),s(x),s(y)) -> 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 + 1 le#!_A(x1,x2) = x1 + x2 + 5 s!_A(x1) = 4 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{4, x1 + 1} le#!_A(x1,x2) = max{5, x1 - 5, x2 - 5} 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#(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: mod(|0|(),y) -> |0|() r7: mod(s(x),|0|()) -> |0|() r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) r10: if_mod(false(),s(x),s(y)) -> 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) = x1 + x2 s_A(x1) = x1 + 2 minus#!_A(x1,x2) = x1 + x2 s!_A(x1) = 3 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: minus#_A(x1,x2) = max{x1, x2} s_A(x1) = x1 + 31 minus#!_A(x1,x2) = max{1, x1 - 32, x2 - 23} s!_A(x1) = 2 partial status: pi(minus#) = [] pi(s) = [] The following rules are removed: p1 Then no dependency pair remains.