YES We show the termination of the TRS R: minus(x,|0|()) -> x minus(s(x),s(y)) -> minus(x,y) double(|0|()) -> |0|() double(s(x)) -> s(s(double(x))) plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) plus(zero(),y) -> y plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) id(x) -> x if(true(),x,y) -> x if(false(),x,y) -> y not(x) -> if(x,false(),true()) gt(s(x),zero()) -> true() gt(zero(),y) -> false() gt(s(x),s(y)) -> gt(x,y) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: minus#(s(x),s(y)) -> minus#(x,y) p2: double#(s(x)) -> double#(x) p3: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))) p4: plus#(s(x),s(y)) -> if#(gt(x,y),x,y) p5: plus#(s(x),s(y)) -> gt#(x,y) p6: plus#(s(x),s(y)) -> if#(not(gt(x,y)),id(x),id(y)) p7: plus#(s(x),s(y)) -> not#(gt(x,y)) p8: plus#(s(x),s(y)) -> id#(x) p9: plus#(s(x),s(y)) -> id#(y) p10: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x)) p11: plus#(s(x),x) -> if#(gt(x,x),id(x),id(x)) p12: plus#(s(x),x) -> gt#(x,x) p13: plus#(s(x),x) -> id#(x) p14: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y))) p15: plus#(id(x),s(y)) -> if#(gt(s(y),y),y,s(y)) p16: plus#(id(x),s(y)) -> gt#(s(y),y) p17: not#(x) -> if#(x,false(),true()) p18: gt#(s(x),s(y)) -> gt#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r7: plus(zero(),y) -> y r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r9: id(x) -> x r10: if(true(),x,y) -> x r11: if(false(),x,y) -> y r12: not(x) -> if(x,false(),true()) r13: gt(s(x),zero()) -> true() r14: gt(zero(),y) -> false() r15: gt(s(x),s(y)) -> gt(x,y) The estimated dependency graph contains the following SCCs: {p1} {p2} {p3, p10, p14} {p18} -- 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: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r7: plus(zero(),y) -> y r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r9: id(x) -> x r10: if(true(),x,y) -> x r11: if(false(),x,y) -> y r12: not(x) -> if(x,false(),true()) r13: gt(s(x),zero()) -> true() r14: gt(zero(),y) -> false() r15: gt(s(x),s(y)) -> gt(x,y) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: minus#_A(x1,x2) = 0 s_A(x1) = max{21, x1 - 23} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: minus#_A(x1,x2) = max{0, x1 - 50, x2 - 50} s_A(x1) = max{105, x1 + 49} minus#!_A(x1,x2) = max{5, x1 - 99, x2 - 124} s!_A(x1) = max{4, x1 - 123} partial status: pi(minus#) = [] 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: double#(s(x)) -> double#(x) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r7: plus(zero(),y) -> y r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r9: id(x) -> x r10: if(true(),x,y) -> x r11: if(false(),x,y) -> y r12: not(x) -> if(x,false(),true()) r13: gt(s(x),zero()) -> true() r14: gt(zero(),y) -> false() r15: gt(s(x),s(y)) -> gt(x,y) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: double#_A(x1) = max{1, x1 - 13} s_A(x1) = max{20, x1 + 12} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: double#_A(x1) = max{0, x1 - 93} s_A(x1) = x1 + 92 double#!_A(x1) = max{32, x1 - 93} s!_A(x1) = max{125, x1 + 31} partial status: pi(double#) = [] 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: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))) p2: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y))) p3: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x)) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r7: plus(zero(),y) -> y r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r9: id(x) -> x r10: if(true(),x,y) -> x r11: if(false(),x,y) -> y r12: not(x) -> if(x,false(),true()) r13: gt(s(x),zero()) -> true() r14: gt(zero(),y) -> false() r15: gt(s(x),s(y)) -> gt(x,y) The set of usable rules consists of r9, r10, r11, r12, r13, r14, r15 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: plus#_A(x1,x2) = max{6, x1 - 253, x2 - 261} s_A(x1) = max{260, x1 + 23} if_A(x1,x2,x3) = max{60, x1 - 84, x2 + 9, x3} gt_A(x1,x2) = max{98, x1 - 53, x2 - 282} not_A(x1) = max{60, x1 - 84} id_A(x1) = max{225, x1 + 5} true_A = 24 false_A = 51 zero_A = 52 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: plus#_A(x1,x2) = max{0, x1 - 19, x2 - 19} s_A(x1) = x1 + 2 if_A(x1,x2,x3) = max{x1 - 14, x2, x3} gt_A(x1,x2) = max{1, x1 - 5, x2 - 3} not_A(x1) = max{17, x1 - 4} id_A(x1) = max{18, x1} true_A = 1 false_A = 1 zero_A = 2 plus#!_A(x1,x2) = max{0, x1 - 45, x2 - 19} s!_A(x1) = 2 if!_A(x1,x2,x3) = max{4, x1 - 3, x2 - 60, x3 - 13} gt!_A(x1,x2) = max{4, x1 - 12, x2 - 3} not!_A(x1) = max{5, x1 - 2} id!_A(x1) = 2 true!_A = 1 false!_A = 4 zero!_A = 5 partial status: pi(plus#) = [] pi(s) = [1] pi(if) = [3] pi(gt) = [] pi(not) = [] pi(id) = [] pi(true) = [] pi(false) = [] pi(zero) = [] The following rules are removed: p1, p3 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y))) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r7: plus(zero(),y) -> y r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r9: id(x) -> x r10: if(true(),x,y) -> x r11: if(false(),x,y) -> y r12: not(x) -> if(x,false(),true()) r13: gt(s(x),zero()) -> true() r14: gt(zero(),y) -> false() r15: gt(s(x),s(y)) -> gt(x,y) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y))) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r7: plus(zero(),y) -> y r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r9: id(x) -> x r10: if(true(),x,y) -> x r11: if(false(),x,y) -> y r12: not(x) -> if(x,false(),true()) r13: gt(s(x),zero()) -> true() r14: gt(zero(),y) -> false() r15: gt(s(x),s(y)) -> gt(x,y) The set of usable rules consists of r10, r11, r13, r14, r15 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: plus#_A(x1,x2) = 1 id_A(x1) = max{0, x1 - 17} s_A(x1) = max{4, x1} if_A(x1,x2,x3) = max{x1 - 2, x2 + 14, x3 + 8} gt_A(x1,x2) = max{13, x1 - 5, x2 + 1} true_A = 0 false_A = 0 zero_A = 1 base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: plus#_A(x1,x2) = max{0, x1 - 88, x2 - 69} id_A(x1) = x1 + 8 s_A(x1) = max{34, x1 + 10} if_A(x1,x2,x3) = max{68, x1 - 1, x2 + 10, x3} gt_A(x1,x2) = max{11, x1 - 43, x2 - 59} true_A = 11 false_A = 11 zero_A = 0 plus#!_A(x1,x2) = max{1, x1 - 4} id!_A(x1) = max{0, x1 - 58} s!_A(x1) = max{5, x1 - 70} if!_A(x1,x2,x3) = max{6, x1 - 129, x2 - 107, x3 - 101} gt!_A(x1,x2) = max{6, x1 - 127, x2 - 60} true!_A = 0 false!_A = 0 zero!_A = 1 partial status: pi(plus#) = [] pi(id) = [] pi(s) = [1] pi(if) = [2, 3] pi(gt) = [2] pi(true) = [] pi(false) = [] pi(zero) = [] 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: gt#(s(x),s(y)) -> gt#(x,y) and R consists of: r1: minus(x,|0|()) -> x r2: minus(s(x),s(y)) -> minus(x,y) r3: double(|0|()) -> |0|() r4: double(s(x)) -> s(s(double(x))) r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r7: plus(zero(),y) -> y r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r9: id(x) -> x r10: if(true(),x,y) -> x r11: if(false(),x,y) -> y r12: not(x) -> if(x,false(),true()) r13: gt(s(x),zero()) -> true() r14: gt(zero(),y) -> false() r15: gt(s(x),s(y)) -> gt(x,y) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: gt#_A(x1,x2) = 0 s_A(x1) = max{5, x1 - 7} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: gt#_A(x1,x2) = max{0, x1 - 89, x2 - 89} s_A(x1) = x1 + 88 gt#!_A(x1,x2) = max{5, x1 - 82, x2 - 89} s!_A(x1) = max{4, x1 - 93} partial status: pi(gt#) = [] pi(s) = [] The following rules are removed: p1 Then no dependency pair remains.