YES 1 decompositions #0 ----------- 1: f(f(f(b()))) -> b() 2: h(h(a(),a()),h(f(f(a())),h(c(),b()))) -> f(c()) @Mutually Orthogonal --- R 1: f(f(f(b()))) -> b() 2: h(h(a(),a()),h(f(f(a())),h(c(),b()))) -> f(c()) --- S 1: f(f(f(b()))) -> b() 2: h(h(a(),a()),h(f(f(a())),h(c(),b()))) -> f(c())