YES

1 decompositions
 #1 -----------
   1: h(a(),a()) -> f(c())
   3: b() -> h(c(),a())

@Knuth and Bendix' criterion
--- R
   1: h(a(),a()) -> f(c())
   3: b() -> h(c(),a())

--- S
   1: h(a(),a()) -> f(c())
   3: b() -> h(c(),a())


NOTE: input TRS is reduced

original is
   1: h(a(),a()) -> f(c())
   2: a() -> a()
   2: a() -> a()
   3: b() -> h(c(),a())

reduced to
   1: h(a(),a()) -> f(c())
   3: b() -> h(c(),a())