YES
Input TRS:
AC symbols: f
      1: f(g(f(h(a()),a())),a()) -> f(h(a()),f(a(),a()))
      2: f(h(a()),g(a())) -> f(g(h(a())),a())
      3: f(g(h(a())),f(b(),f(b(),b()))) -> f(g(f(h(a()),a())),a())
      4: f(h(a()),a()) -> f(h(a()),b())
Number of strict rules: 4
Direct POLO(bPol) ... removes: 4 1 3
      h 	w: x1 + 8365
      a 	w: 1
      b 	w: 0
      f 	w: x1 + x2 + 3
      g 	w: x1 + 1
Number of strict rules: 1
Direct POLO(bPol) ... failed.
Uncurrying ... failed.
Dependency Pairs:
     #1: #f(h(a()),g(a())) -> #f(g(h(a())),a())
     #2: #f(x,f(y,z)) ->= #f(f(x,y),z)
     #3: #f(x,f(y,z)) ->= #f(x,y)
Number of SCCs: 1, DPs: 3
  SCC { #1..3 }
POLO(Sum)... succeeded.
      h 	w: x1 + 1
      a 	w: 1
      b 	w: 0
      f 	w: x1 + x2 + 1
      #f 	w: x1 + x2
      g 	w: 11797
    USABLE RULES: { 2 5 }
    Removed DPs: #1 #3
Number of SCCs: 1, DPs: 1
  SCC { #2 }
only weak rules.
Number of SCCs: 0, DPs: 0
Next Dependency Pairs:
     #4: #f(f(h(a()),g(a())),_1) -> #f(f(g(h(a())),a()),_1)
     #5: #f(x,f(y,z)) ->= #f(f(x,y),z)
     #6: #f(x,f(y,z)) ->= #f(x,y)
Number of SCCs: 1, DPs: 3
  SCC { #4..6 }
POLO(Sum)... succeeded.
      h 	w: x1 + 3433
      a 	w: 1
      b 	w: 0
      f 	w: x1 + x2 + 1
      #f 	w: x1 + x2
      g 	w: 8365
    USABLE RULES: { 2 5 }
    Removed DPs: #4 #6
Number of SCCs: 1, DPs: 1
  SCC { #5 }
only weak rules.
Number of SCCs: 0, DPs: 0