YES

1 decompositions
 #0 -----------
   1: a(a(b(a(b(a(b(a(b(x))))))))) -> a(b(a(b(a(b(a(b(a(a(a(a(a(b(x))))))))))))))

@Mutually Orthogonal
--- R
   1: a(a(b(a(b(a(b(a(b(x))))))))) -> a(b(a(b(a(b(a(b(a(a(a(a(a(b(x))))))))))))))

--- S
   1: a(a(b(a(b(a(b(a(b(x))))))))) -> a(b(a(b(a(b(a(b(a(a(a(a(a(b(x))))))))))))))