YES

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

@Strongly Commuting
--- R
   1: b(a(b(b(x)))) -> b(b(b(a(b(x)))))
   2: b(a(a(b(b(x))))) -> b(a(b(b(a(a(b(x)))))))
   3: b(a(a(a(b(b(x)))))) -> b(a(a(b(b(a(a(a(b(x)))))))))

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