YES

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

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

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