YES We show the termination of the TRS R: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z))) times(x,|0|()) -> |0|() times(x,s(y)) -> plus(times(x,y),x) plus(x,|0|()) -> x plus(x,s(y)) -> s(plus(x,y)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: times#(x,plus(y,s(z))) -> plus#(times(x,plus(y,times(s(z),|0|()))),times(x,s(z))) p2: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|()))) p3: times#(x,plus(y,s(z))) -> plus#(y,times(s(z),|0|())) p4: times#(x,plus(y,s(z))) -> times#(s(z),|0|()) p5: times#(x,plus(y,s(z))) -> times#(x,s(z)) p6: times#(x,s(y)) -> plus#(times(x,y),x) p7: times#(x,s(y)) -> times#(x,y) p8: plus#(x,s(y)) -> plus#(x,y) and R consists of: r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z))) r2: times(x,|0|()) -> |0|() r3: times(x,s(y)) -> plus(times(x,y),x) r4: plus(x,|0|()) -> x r5: plus(x,s(y)) -> s(plus(x,y)) The estimated dependency graph contains the following SCCs: {p2, p5, p7} {p8} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: times#(x,s(y)) -> times#(x,y) p2: times#(x,plus(y,s(z))) -> times#(x,s(z)) p3: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|()))) and R consists of: r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z))) r2: times(x,|0|()) -> |0|() r3: times(x,s(y)) -> plus(times(x,y),x) r4: plus(x,|0|()) -> x r5: plus(x,s(y)) -> s(plus(x,y)) The set of usable rules consists of r2, 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: times#_A(x1,x2) = x2 s_A(x1) = x1 + 1 plus_A(x1,x2) = x1 + x2 + 2 times_A(x1,x2) = x1 |0|_A() = 0 times#!_A(x1,x2) = x2 + 2 s!_A(x1) = 1 plus!_A(x1,x2) = 2 times!_A(x1,x2) = x1 + 3 |0|!_A() = 0 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: times#_A(x1,x2) = max{0, x1 - 101, x2 - 52} s_A(x1) = max{26, x1} plus_A(x1,x2) = max{x1 + 22, x2 + 13} times_A(x1,x2) = max{12, x2 - 73} |0|_A = 12 times#!_A(x1,x2) = max{0, x1 - 1} s!_A(x1) = max{64, x1 - 25} plus!_A(x1,x2) = max{65, x1 + 50, x2 - 6} times!_A(x1,x2) = max{63, x1 - 52} |0|!_A = 63 partial status: pi(times#) = [] pi(s) = [1] pi(plus) = [1, 2] pi(times) = [2] pi(|0|) = [] The following rules are removed: p1, p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|()))) and R consists of: r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z))) r2: times(x,|0|()) -> |0|() r3: times(x,s(y)) -> plus(times(x,y),x) r4: plus(x,|0|()) -> x r5: plus(x,s(y)) -> s(plus(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: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|()))) and R consists of: r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z))) r2: times(x,|0|()) -> |0|() r3: times(x,s(y)) -> plus(times(x,y),x) r4: plus(x,|0|()) -> x r5: plus(x,s(y)) -> s(plus(x,y)) The set of usable rules consists of r2, 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: times#_A(x1,x2) = x2 plus_A(x1,x2) = x1 + x2 + 1 s_A(x1) = x1 + 1 times_A(x1,x2) = x1 |0|_A() = 0 times#!_A(x1,x2) = 3 plus!_A(x1,x2) = 4 s!_A(x1) = 2 times!_A(x1,x2) = 1 |0|!_A() = 0 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: times#_A(x1,x2) = max{42, x1 - 97, x2 - 62} plus_A(x1,x2) = max{61, x1 + 5, x2 + 44} s_A(x1) = 21 times_A(x1,x2) = max{60, x1 - 99, x2 - 18} |0|_A = 40 times#!_A(x1,x2) = max{0, x2 - 20} plus!_A(x1,x2) = max{85, x1 - 49, x2 - 72} s!_A(x1) = x1 + 23 times!_A(x1,x2) = max{1, x2 - 18} |0|!_A = 44 partial status: pi(times#) = [2] pi(plus) = [1, 2] pi(s) = [] pi(times) = [2] pi(|0|) = [] 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#(x,s(y)) -> plus#(x,y) and R consists of: r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z))) r2: times(x,|0|()) -> |0|() r3: times(x,s(y)) -> plus(times(x,y),x) r4: plus(x,|0|()) -> x r5: plus(x,s(y)) -> s(plus(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!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: plus#_A(x1,x2) = x2 s_A(x1) = x1 + 2 plus#!_A(x1,x2) = x2 + 3 s!_A(x1) = 2 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: plus#_A(x1,x2) = 0 s_A(x1) = max{4, x1 - 16} plus#!_A(x1,x2) = max{25, x1 - 1, x2 - 1} s!_A(x1) = 24 partial status: pi(plus#) = [1] pi(s) = [1] The following rules are removed: p1 Then no dependency pair remains.