YES
Time: 0.013

Problem:
 Equations:
  plusAC(plusAC(x2,x3),x4) -> plusAC(x2,plusAC(x3,x4))
  plusAC(x2,x3) -> plusAC(x3,x2)
  plusAC(x2,plusAC(x3,x4)) -> plusAC(plusAC(x2,x3),x4)
  plusAC(x3,x2) -> plusAC(x2,x3)
 TRS:
  f(plusAC(x,y)) -> plusAC(f(x),f(y))
  plusAC(x,0()) -> 0()
  f(0()) -> 0()

Proof:
 AC-RPO Processor:
  precedence:
   0 > f > plusAC
  status:
   
  problem:
   Equations:
    plusAC(plusAC(x2,x3),x4) -> plusAC(x2,plusAC(x3,x4))
    plusAC(x2,x3) -> plusAC(x3,x2)
    plusAC(x2,plusAC(x3,x4)) -> plusAC(plusAC(x2,x3),x4)
    plusAC(x3,x2) -> plusAC(x2,x3)
   TRS:
    
  Qed