YES

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

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

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