YES We show the termination of the TRS R: f(s(x),y) -> f(x,s(x)) f(x,s(y)) -> f(y,x) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: f#(s(x),y) -> f#(x,s(x)) p2: f#(x,s(y)) -> f#(y,x) and R consists of: r1: f(s(x),y) -> f(x,s(x)) r2: f(x,s(y)) -> f(y,x) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: f#(s(x),y) -> f#(x,s(x)) p2: f#(x,s(y)) -> f#(y,x) and R consists of: r1: f(s(x),y) -> f(x,s(x)) r2: f(x,s(y)) -> f(y,x) 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: f#_A(x1,x2) = max{0, x1 - 10, x2 - 14} s_A(x1) = max{9, x1 + 5} f#!_A(x1,x2) = max{2, x1 - 6, x2 - 11} s!_A(x1) = max{1, x1 - 12} 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: f#_A(x1,x2) = max{0, x1 - 3, x2 - 3} s_A(x1) = max{2, x1} f#!_A(x1,x2) = max{1, x1 - 3, x2 - 3} s!_A(x1) = 2 partial status: pi(f#) = [] pi(s) = [1] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: f#(x,s(y)) -> f#(y,x) and R consists of: r1: f(s(x),y) -> f(x,s(x)) r2: f(x,s(y)) -> f(y,x) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: f#(x,s(y)) -> f#(y,x) and R consists of: r1: f(s(x),y) -> f(x,s(x)) r2: f(x,s(y)) -> f(y,x) 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: f#_A(x1,x2) = 0 s_A(x1) = x1 + 2 f#!_A(x1,x2) = max{x1, x2 - 1} s!_A(x1) = max{0, x1 - 3} 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: f#_A(x1,x2) = 0 s_A(x1) = max{1, x1 - 2} f#!_A(x1,x2) = max{1, x2 - 2} s!_A(x1) = max{0, x1 - 6} partial status: pi(f#) = [] pi(s) = [1] The following rules are removed: p1 Then no dependency pair remains.