YES We show the termination of R/S, where R is not(true) -> false not(false) -> true evenodd(x,0) -> not(evenodd(x,s(0))) evenodd(0,s(0)) -> false evenodd(s(x),s(0)) -> evenodd(x,0) and S is: rand(x) -> x rand(x) -> rand(s(x)) 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 evenodd#(x,0) -> not#(evenodd(x,s(0))) evenodd#(x,0) -> evenodd#(x,s(0)) evenodd#(s(x),s(0)) -> evenodd#(x,0) and Q consists of the rules: not(true) -> false not(false) -> true evenodd(x,0) -> not(evenodd(x,s(0))) evenodd(0,s(0)) -> false evenodd(s(x),s(0)) -> evenodd(x,0) rand(x) -> x rand(x) -> rand(s(x)) The weakly monotone algebra (N^3, >_lex) with evenodd#_A((x1,y1,z1),(x2,y2,z2)) = (3, 1, 1) 0_A = (1, 1, 0) not#_A((x1,y1,z1)) = (x1, 0, 0) evenodd_A((x1,y1,z1),(x2,y2,z2)) = (2, 3, 1) s_A((x1,y1,z1)) = (0, 0, 1) not_A((x1,y1,z1)) = (x1, 2, 1) true_A = (1, 1, 1) false_A = (1, 1, 1) rand_A((x1,y1,z1)) = (x1 + 1, y1 + 1, 0) strictly orients the following dependency pairs: evenodd#(x,0) -> not#(evenodd(x,s(0))) We remove them from the set of dependency pairs. The weakly monotone algebra (N^3, >_lex) with evenodd#_A((x1,y1,z1),(x2,y2,z2)) = (x1, y1, z1) 0_A = (1, 1, 1) s_A((x1,y1,z1)) = (x1, y1 + 1, z1 + 1) not_A((x1,y1,z1)) = (0, 2, 1) true_A = (0, 1, 1) false_A = (0, 1, 1) evenodd_A((x1,y1,z1),(x2,y2,z2)) = (x1, 3, 1) rand_A((x1,y1,z1)) = (x1 + 1, 1, 0) strictly orients the following dependency pairs: evenodd#(s(x),s(0)) -> evenodd#(x,0) We remove them from the set of dependency pairs. The weakly monotone algebra (N^3, >_lex) with evenodd#_A((x1,y1,z1),(x2,y2,z2)) = (x1 + x2, y1 + y2, z1 + z2) 0_A = (1, 1, 2) s_A((x1,y1,z1)) = (0, 2, 1) not_A((x1,y1,z1)) = (2, 3, 2) true_A = (1, 1, 1) false_A = (1, 2, 2) evenodd_A((x1,y1,z1),(x2,y2,z2)) = (3, 1, 1) rand_A((x1,y1,z1)) = (x1 + 1, 1, 0) strictly orients the following dependency pairs: evenodd#(x,0) -> evenodd#(x,s(0)) We remove them from the set of dependency pairs. No dependency pair remains.