YES

1 decompositions
 #1 -----------
   1: a(c(x1)) -> b(b(x1))
   2: a(c(x2)) -> b(a(x2))
   3: b(a(x3)) -> c(c(x3))
   5: b(b(x5)) -> b(c(x5))
   6: b(c(x6)) -> c(b(x6))
   7: c(c(x7)) -> b(c(x7))

@Rule Labeling
--- R
   1: a(c(x1)) -> b(b(x1))
   2: a(c(x2)) -> b(a(x2))
   3: b(a(x3)) -> c(c(x3))
   5: b(b(x5)) -> b(c(x5))
   6: b(c(x6)) -> c(b(x6))
   7: c(c(x7)) -> b(c(x7))

--- S
   1: a(c(x1)) -> b(b(x1))
   2: a(c(x2)) -> b(a(x2))
   3: b(a(x3)) -> c(c(x3))
   5: b(b(x5)) -> b(c(x5))
   6: b(c(x6)) -> c(b(x6))
   7: c(c(x7)) -> b(c(x7))


NOTE: input TRS is reduced

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

reduced to
   1: a(c(x1)) -> b(b(x1))
   2: a(c(x2)) -> b(a(x2))
   3: b(a(x3)) -> c(c(x3))
   5: b(b(x5)) -> b(c(x5))
   6: b(c(x6)) -> c(b(x6))
   7: c(c(x7)) -> b(c(x7))