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: max/plus interpretation on natural numbers: if_mod#_A(x1,x2,x3) = max{1, x1 - 4, x2 - 1} true_A = 5 s_A(x1) = x1 + 8 mod#_A(x1,x2) = max{7, x1 + 2} minus_A(x1,x2) = x1 + 5 le_A(x1,x2) = x2 + 7 |0|_A = 0 false_A = 1 base order 2: 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 + 9 le_A(x1,x2) = 1 |0|_A() = 2 false_A() = 1 if_mod#!_A(x1,x2,x3) = 7 true!_A() = 9 s!_A(x1) = 8 mod#!_A(x1,x2) = 6 minus!_A(x1,x2) = x2 + 6 le!_A(x1,x2) = x1 + x2 + 3 |0|!_A() = 6 false!_A() = 7 partial status: pi(if_mod#) = [] pi(true) = [] pi(s) = [1] pi(mod#) = [1] pi(minus) = [] pi(le) = [2] 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 r1, r2, r3, r4, r5, r6, r7, r10 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: le#_A(x1,x2) = max{9, x1 - 9, x2 - 9} s_A(x1) = max{8, x1 + 3} le_A(x1,x2) = max{12, x1 - 9, x2 - 9} |0|_A = 7 true_A = 1 false_A = 4 minus_A(x1,x2) = max{15, x1 + 1} mod_A(x1,x2) = max{13, x1 + 3, x2 - 10} if_mod_A(x1,x2,x3) = max{14, x1 - 15, x2 + 2, x3 - 9} base order 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 + 2 le_A(x1,x2) = x1 + x2 + 1 |0|_A() = 1 true_A() = 1 false_A() = 1 minus_A(x1,x2) = x1 + x2 + 1 mod_A(x1,x2) = x1 if_mod_A(x1,x2,x3) = x2 le#!_A(x1,x2) = x1 + x2 + 6 s!_A(x1) = x1 + 5 le!_A(x1,x2) = x1 + 5 |0|!_A() = 1 true!_A() = 0 false!_A() = 0 minus!_A(x1,x2) = x1 + 4 mod!_A(x1,x2) = x1 + x2 + 1 if_mod!_A(x1,x2,x3) = x2 + x3 + 1 partial status: pi(le#) = [] pi(s) = [1] pi(le) = [] pi(|0|) = [] pi(true) = [] pi(false) = [] pi(minus) = [] pi(mod) = [1] pi(if_mod) = [2] 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 r1, r2, r3, r4, r5, r6, r7, r10 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: minus#_A(x1,x2) = max{0, x1 - 9} s_A(x1) = max{8, x1 + 3} le_A(x1,x2) = max{12, x1 - 9, x2 - 9} |0|_A = 7 true_A = 1 false_A = 4 minus_A(x1,x2) = max{15, x1 + 1} mod_A(x1,x2) = max{13, x1 + 3, x2 - 10} if_mod_A(x1,x2,x3) = max{14, x1 - 15, x2 + 2, x3 - 9} base order 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) = x1 + x2 s_A(x1) = x1 + 2 le_A(x1,x2) = x1 + x2 + 1 |0|_A() = 1 true_A() = 1 false_A() = 1 minus_A(x1,x2) = x1 + x2 + 1 mod_A(x1,x2) = x1 if_mod_A(x1,x2,x3) = x2 minus#!_A(x1,x2) = x1 + x2 + 6 s!_A(x1) = x1 + 5 le!_A(x1,x2) = x1 + 5 |0|!_A() = 1 true!_A() = 0 false!_A() = 0 minus!_A(x1,x2) = x1 + 4 mod!_A(x1,x2) = x1 + x2 + 1 if_mod!_A(x1,x2,x3) = x2 + x3 + 1 partial status: pi(minus#) = [] pi(s) = [1] pi(le) = [] pi(|0|) = [] pi(true) = [] pi(false) = [] pi(minus) = [] pi(mod) = [1] pi(if_mod) = [2] The following rules are removed: p1 Then no dependency pair remains.