YES We show the termination of R/S, where R is f(a,g(y),z) -> f(a,y,g(y)) f(b,g(y),z) -> f(a,y,z) a -> b and S is: f(x,y,z) -> f(x,y,g(z)) Since R almost dominates S and S is non-duplicating, it is enough to show finiteness of (P, Q). Here P consists of the dependency pairs f#(a,g(y),z) -> f#(a,y,g(y)) f#(a,g(y),z) -> a# f#(b,g(y),z) -> f#(a,y,z) f#(b,g(y),z) -> a# and Q consists of the rules: f(a,g(y),z) -> f(a,y,g(y)) f(b,g(y),z) -> f(a,y,z) a -> b f(x,y,z) -> f(x,y,g(z)) f#(x,y,z) -> f#(x,y,g(z)) The weakly monotone algebra (N, >) with f#_A(x1,x2,x3) = x2 a_A = 1 g_A(x1) = x1 + 1 a#_A = 0 b_A = 1 f_A(x1,x2,x3) = x2 strictly orients the following dependency pairs: f#(a,g(y),z) -> f#(a,y,g(y)) f#(a,g(y),z) -> a# f#(b,g(y),z) -> f#(a,y,z) f#(b,g(y),z) -> a# We remove them from the set of dependency pairs. No dependency pair remains.