YES We show the termination of the TRS R: f(c(s(x),y)) -> f(c(x,s(y))) g(c(x,s(y))) -> g(c(s(x),y)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: f#(c(s(x),y)) -> f#(c(x,s(y))) p2: g#(c(x,s(y))) -> g#(c(s(x),y)) and R consists of: r1: f(c(s(x),y)) -> f(c(x,s(y))) r2: g(c(x,s(y))) -> g(c(s(x),y)) 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#(c(s(x),y)) -> f#(c(x,s(y))) and R consists of: r1: f(c(s(x),y)) -> f(c(x,s(y))) r2: g(c(x,s(y))) -> g(c(s(x),y)) The set of usable rules consists of (no rules) Take the reduction pair: lexicographic combination of reduction pairs: 1. matrix interpretations: carrier: N^1 order: standard order interpretations: f#_A(x1) = x1 c_A(x1,x2) = x1 s_A(x1) = x1 + 1 2. matrix interpretations: carrier: N^1 order: standard order interpretations: f#_A(x1) = x1 c_A(x1,x2) = 0 s_A(x1) = x1 + 1 The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: g#(c(x,s(y))) -> g#(c(s(x),y)) and R consists of: r1: f(c(s(x),y)) -> f(c(x,s(y))) r2: g(c(x,s(y))) -> g(c(s(x),y)) The set of usable rules consists of (no rules) Take the reduction pair: lexicographic combination of reduction pairs: 1. matrix interpretations: carrier: N^1 order: standard order interpretations: g#_A(x1) = x1 c_A(x1,x2) = x2 s_A(x1) = x1 + 1 2. matrix interpretations: carrier: N^1 order: standard order interpretations: g#_A(x1) = x1 c_A(x1,x2) = x2 s_A(x1) = x1 + 1 The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains.