YES
Time: 0.019

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-KBO Processor:
  precedence:
   f > 0 ~ plusAC
   weight function:
    [0] = 4,
    
    [f](x0) = x0,
    
    [plusAC](x0, x1) = x0 + x1 + 1
   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