YES

1 decompositions
 #1 -----------
   1: a(c(x1)) -> b(a(x1))
   3: a(a(x3)) -> a(b(x3))
   4: b(b(x4)) -> a(c(x4))
   5: c(c(x5)) -> c(a(x5))
   6: c(b(x6)) -> a(c(x6))
   7: a(b(x7)) -> a(c(x7))

@Rule Labeling
--- R
   1: a(c(x1)) -> b(a(x1))
   3: a(a(x3)) -> a(b(x3))
   4: b(b(x4)) -> a(c(x4))
   5: c(c(x5)) -> c(a(x5))
   6: c(b(x6)) -> a(c(x6))
   7: a(b(x7)) -> a(c(x7))

--- S
   1: a(c(x1)) -> b(a(x1))
   3: a(a(x3)) -> a(b(x3))
   4: b(b(x4)) -> a(c(x4))
   5: c(c(x5)) -> c(a(x5))
   6: c(b(x6)) -> a(c(x6))
   7: a(b(x7)) -> a(c(x7))


NOTE: input TRS is reduced

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

reduced to
   1: a(c(x1)) -> b(a(x1))
   3: a(a(x3)) -> a(b(x3))
   4: b(b(x4)) -> a(c(x4))
   5: c(c(x5)) -> c(a(x5))
   6: c(b(x6)) -> a(c(x6))
   7: a(b(x7)) -> a(c(x7))