YES We show the termination of the TRS R: f(X) -> g(n__h(n__f(X))) h(X) -> n__h(X) f(X) -> n__f(X) activate(n__h(X)) -> h(activate(X)) activate(n__f(X)) -> f(activate(X)) activate(X) -> X -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: activate#(n__h(X)) -> h#(activate(X)) p2: activate#(n__h(X)) -> activate#(X) p3: activate#(n__f(X)) -> f#(activate(X)) p4: activate#(n__f(X)) -> activate#(X) and R consists of: r1: f(X) -> g(n__h(n__f(X))) r2: h(X) -> n__h(X) r3: f(X) -> n__f(X) r4: activate(n__h(X)) -> h(activate(X)) r5: activate(n__f(X)) -> f(activate(X)) r6: activate(X) -> X The estimated dependency graph contains the following SCCs: {p2, p4} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: activate#(n__h(X)) -> activate#(X) p2: activate#(n__f(X)) -> activate#(X) and R consists of: r1: f(X) -> g(n__h(n__f(X))) r2: h(X) -> n__h(X) r3: f(X) -> n__f(X) r4: activate(n__h(X)) -> h(activate(X)) r5: activate(n__f(X)) -> f(activate(X)) r6: activate(X) -> X The set of usable rules consists of (no rules) Take the reduction pair: lexicographic combination of reduction pairs: 1. matrix interpretations: carrier: N^2 order: standard order interpretations: activate#_A(x1) = ((1,1),(1,0)) x1 n__h_A(x1) = ((1,1),(1,1)) x1 + (1,1) n__f_A(x1) = ((1,1),(1,1)) x1 + (1,1) 2. matrix interpretations: carrier: N^2 order: standard order interpretations: activate#_A(x1) = ((1,1),(1,0)) x1 n__h_A(x1) = ((1,1),(0,1)) x1 + (1,1) n__f_A(x1) = ((1,1),(1,1)) x1 + (1,1) 3. matrix interpretations: carrier: N^2 order: standard order interpretations: activate#_A(x1) = ((1,1),(1,1)) x1 n__h_A(x1) = ((0,1),(1,1)) x1 + (1,1) n__f_A(x1) = ((1,1),(1,1)) x1 + (1,1) The next rules are strictly ordered: p1, p2 We remove them from the problem. Then no dependency pair remains.