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(|0|(),y) -> |0|() minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) if_minus(true(),s(x),y) -> |0|() if_minus(false(),s(x),y) -> s(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),y) -> if_minus#(le(s(x),y),s(x),y) p3: minus#(s(x),y) -> le#(s(x),y) p4: if_minus#(false(),s(x),y) -> minus#(x,y) p5: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y)) p6: mod#(s(x),s(y)) -> le#(y,x) p7: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y)) p8: 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(|0|(),y) -> |0|() r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) r6: if_minus(true(),s(x),y) -> |0|() r7: if_minus(false(),s(x),y) -> s(minus(x,y)) r8: mod(|0|(),y) -> |0|() r9: mod(s(x),|0|()) -> |0|() r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) r12: if_mod(false(),s(x),s(y)) -> s(x) The estimated dependency graph contains the following SCCs: {p5, p7} {p2, p4} {p1} -- 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(|0|(),y) -> |0|() r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) r6: if_minus(true(),s(x),y) -> |0|() r7: if_minus(false(),s(x),y) -> s(minus(x,y)) r8: mod(|0|(),y) -> |0|() r9: mod(s(x),|0|()) -> |0|() r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) r12: if_mod(false(),s(x),s(y)) -> s(x) The set of usable rules consists of r4, r5, r6, r7 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{0, x2 - 20} true_A = 74 s_A(x1) = x1 + 39 mod#_A(x1,x2) = max{1, x1 - 19} minus_A(x1,x2) = x1 + 37 le_A(x1,x2) = max{2, x1 - 40} |0|_A = 75 if_minus_A(x1,x2,x3) = max{56, x2 + 37} false_A = 17 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() = 3 s_A(x1) = x1 + 5 mod#_A(x1,x2) = x2 minus_A(x1,x2) = x1 + 1 le_A(x1,x2) = x1 + 1 |0|_A() = 1 if_minus_A(x1,x2,x3) = x2 + 1 false_A() = 7 if_mod#!_A(x1,x2,x3) = 7 true!_A() = 7 s!_A(x1) = x1 + 9 mod#!_A(x1,x2) = x1 + x2 minus!_A(x1,x2) = x1 + x2 + 10 le!_A(x1,x2) = 15 |0|!_A() = 12 if_minus!_A(x1,x2,x3) = x2 + 6 false!_A() = 13 partial status: pi(if_mod#) = [] pi(true) = [] pi(s) = [] pi(mod#) = [] pi(minus) = [] pi(le) = [] pi(|0|) = [] pi(if_minus) = [] 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: if_minus#(false(),s(x),y) -> minus#(x,y) p2: minus#(s(x),y) -> if_minus#(le(s(x),y),s(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(|0|(),y) -> |0|() r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) r6: if_minus(true(),s(x),y) -> |0|() r7: if_minus(false(),s(x),y) -> s(minus(x,y)) r8: mod(|0|(),y) -> |0|() r9: mod(s(x),|0|()) -> |0|() r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) r12: if_mod(false(),s(x),s(y)) -> s(x) The set of usable rules consists of r1, r2, r3, r4, r8 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: if_minus#_A(x1,x2,x3) = max{2, x1 - 18, x2 - 27} false_A = 20 s_A(x1) = max{40, x1 + 14} minus#_A(x1,x2) = max{1, x1 - 14} le_A(x1,x2) = max{25, x1 - 23} |0|_A = 21 true_A = 1 minus_A(x1,x2) = 22 mod_A(x1,x2) = max{41, x1 - 42, x2 - 64} 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_minus#_A(x1,x2,x3) = x2 false_A() = 1 s_A(x1) = x1 + 1 minus#_A(x1,x2) = x1 le_A(x1,x2) = x1 + 1 |0|_A() = 7 true_A() = 1 minus_A(x1,x2) = x1 + x2 + 5 mod_A(x1,x2) = x1 if_minus#!_A(x1,x2,x3) = x2 + x3 + 3 false!_A() = 3 s!_A(x1) = 6 minus#!_A(x1,x2) = x1 + x2 + 4 le!_A(x1,x2) = 4 |0|!_A() = 7 true!_A() = 0 minus!_A(x1,x2) = x2 + 8 mod!_A(x1,x2) = x1 + x2 + 1 partial status: pi(if_minus#) = [] pi(false) = [] pi(s) = [1] pi(minus#) = [] pi(le) = [] pi(|0|) = [] pi(true) = [] pi(minus) = [] pi(mod) = [] 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(|0|(),y) -> |0|() r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y) r6: if_minus(true(),s(x),y) -> |0|() r7: if_minus(false(),s(x),y) -> s(minus(x,y)) r8: mod(|0|(),y) -> |0|() r9: mod(s(x),|0|()) -> |0|() r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) r12: if_mod(false(),s(x),s(y)) -> s(x) The set of usable rules consists of r4, r8 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: le#_A(x1,x2) = max{1, x1 - 2} s_A(x1) = x1 + 14 minus_A(x1,x2) = max{30, x1 + 26, x2 + 26} |0|_A = 12 mod_A(x1,x2) = max{13, x2 - 13} 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 + 1 minus_A(x1,x2) = x1 + x2 + 3 |0|_A() = 4 mod_A(x1,x2) = x1 le#!_A(x1,x2) = 5 s!_A(x1) = 5 minus!_A(x1,x2) = 6 |0|!_A() = 1 mod!_A(x1,x2) = x1 + x2 partial status: pi(le#) = [] pi(s) = [1] pi(minus) = [1, 2] pi(|0|) = [] pi(mod) = [] The following rules are removed: p1 Then no dependency pair remains.