YES We show the termination of the TRS R: times(x,|0|()) -> |0|() times(x,s(y)) -> plus(times(x,y),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: times#(x,s(y)) -> plus#(times(x,y),x) p2: times#(x,s(y)) -> times#(x,y) 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: times(x,|0|()) -> |0|() r2: times(x,s(y)) -> plus(times(x,y),x) r3: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r4: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r5: plus(zero(),y) -> y r6: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r7: id(x) -> x r8: if(true(),x,y) -> x r9: if(false(),x,y) -> y r10: not(x) -> if(x,false(),true()) r11: gt(s(x),zero()) -> true() r12: gt(zero(),y) -> false() r13: gt(s(x),s(y)) -> gt(x,y) The estimated dependency graph contains the following SCCs: {p2} {p3, p10, p14} {p18} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: times#(x,s(y)) -> times#(x,y) and R consists of: r1: times(x,|0|()) -> |0|() r2: times(x,s(y)) -> plus(times(x,y),x) r3: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r4: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r5: plus(zero(),y) -> y r6: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r7: id(x) -> x r8: if(true(),x,y) -> x r9: if(false(),x,y) -> y r10: not(x) -> if(x,false(),true()) r11: gt(s(x),zero()) -> true() r12: gt(zero(),y) -> false() r13: 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: 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: times#_A(x1,x2) = max{0, x2 - 30} s_A(x1) = x1 + 29 times#!_A(x1,x2) = max{1, x2 - 30} s!_A(x1) = max{65, x1 + 32} 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: times#_A(x1,x2) = x2 s_A(x1) = x1 + 2 times#!_A(x1,x2) = x2 s!_A(x1) = x1 + 4 partial status: pi(times#) = [] pi(s) = [1] 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: times(x,|0|()) -> |0|() r2: times(x,s(y)) -> plus(times(x,y),x) r3: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r4: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r5: plus(zero(),y) -> y r6: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r7: id(x) -> x r8: if(true(),x,y) -> x r9: if(false(),x,y) -> y r10: not(x) -> if(x,false(),true()) r11: gt(s(x),zero()) -> true() r12: gt(zero(),y) -> false() r13: gt(s(x),s(y)) -> gt(x,y) The set of usable rules consists of r7, r8, r9 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: plus#_A(x1,x2) = 0 s_A(x1) = max{19, x1 + 15} if_A(x1,x2,x3) = max{18, x2, x3} gt_A(x1,x2) = 1 not_A(x1) = 19 id_A(x1) = max{8, x1} true_A = 20 false_A = 20 plus#!_A(x1,x2) = max{x1 + 12, x2 + 10} s!_A(x1) = max{7, x1 + 3} if!_A(x1,x2,x3) = max{38, x2 + 26, x3 + 28} gt!_A(x1,x2) = max{32, x2 - 20} not!_A(x1) = x1 + 25 id!_A(x1) = 8 true!_A = 33 false!_A = 32 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: plus#_A(x1,x2) = 0 s_A(x1) = 7 if_A(x1,x2,x3) = x2 + x3 + 1 gt_A(x1,x2) = 1 not_A(x1) = 7 id_A(x1) = x1 + 2 true_A() = 2 false_A() = 5 plus#!_A(x1,x2) = x1 s!_A(x1) = 11 if!_A(x1,x2,x3) = x2 + 6 gt!_A(x1,x2) = 1 not!_A(x1) = 10 id!_A(x1) = 4 true!_A() = 12 false!_A() = 0 partial status: pi(plus#) = [] pi(s) = [1] pi(if) = [2, 3] pi(gt) = [] pi(not) = [] pi(id) = [1] pi(true) = [] pi(false) = [] 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: gt#(s(x),s(y)) -> gt#(x,y) and R consists of: r1: times(x,|0|()) -> |0|() r2: times(x,s(y)) -> plus(times(x,y),x) r3: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y))))) r4: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x)) r5: plus(zero(),y) -> y r6: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y)))) r7: id(x) -> x r8: if(true(),x,y) -> x r9: if(false(),x,y) -> y r10: not(x) -> if(x,false(),true()) r11: gt(s(x),zero()) -> true() r12: gt(zero(),y) -> false() r13: 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: 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: gt#_A(x1,x2) = max{0, x1 - 1, x2 - 1} s_A(x1) = x1 gt#!_A(x1,x2) = max{35, x1 - 1, x2 - 1} s!_A(x1) = max{34, x1 + 9} 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: gt#_A(x1,x2) = x1 + x2 s_A(x1) = x1 + 1 gt#!_A(x1,x2) = x1 + x2 + 3 s!_A(x1) = x1 + 2 partial status: pi(gt#) = [] pi(s) = [1] The following rules are removed: p1 Then no dependency pair remains.