YES

1 decompositions
 #1 -----------
   1: c(b(x1)) -> c(a(x1))
   3: b(b(x3)) -> c(a(x3))
   4: a(a(x4)) -> c(a(x4))
   6: b(a(x6)) -> c(b(x6))
   7: c(a(x7)) -> a(c(x7))
   8: c(c(x8)) -> c(a(x8))

@Rule Labeling
--- R
   1: c(b(x1)) -> c(a(x1))
   3: b(b(x3)) -> c(a(x3))
   4: a(a(x4)) -> c(a(x4))
   6: b(a(x6)) -> c(b(x6))
   7: c(a(x7)) -> a(c(x7))
   8: c(c(x8)) -> c(a(x8))

--- S
   1: c(b(x1)) -> c(a(x1))
   3: b(b(x3)) -> c(a(x3))
   4: a(a(x4)) -> c(a(x4))
   6: b(a(x6)) -> c(b(x6))
   7: c(a(x7)) -> a(c(x7))
   8: c(c(x8)) -> c(a(x8))


NOTE: input TRS is reduced

original is
   1: c(b(x1)) -> c(a(x1))
   2: b(b(x2)) -> b(b(x2))
   3: b(b(x3)) -> c(a(x3))
   4: a(a(x4)) -> c(a(x4))
   5: c(c(x5)) -> a(c(x5))
   6: b(a(x6)) -> c(b(x6))
   7: c(a(x7)) -> a(c(x7))
   8: c(c(x8)) -> c(a(x8))

reduced to
   1: c(b(x1)) -> c(a(x1))
   3: b(b(x3)) -> c(a(x3))
   4: a(a(x4)) -> c(a(x4))
   6: b(a(x6)) -> c(b(x6))
   7: c(a(x7)) -> a(c(x7))
   8: c(c(x8)) -> c(a(x8))