YES
Time: 0.061

Problem:
 Equations:
  pAC(pAC(x2,x3),x4) -> pAC(x2,pAC(x3,x4))
  pAC(x2,x3) -> pAC(x3,x2)
  pAC(x2,pAC(x3,x4)) -> pAC(pAC(x2,x3),x4)
  pAC(x3,x2) -> pAC(x2,x3)
 TRS:
  pAC(x,one()) -> x
  pAC(x,x) -> x
  pAC(T(x),x) -> T(x)
  T(pAC(x,T(y))) -> pAC(T(pAC(x,y)),T(y))
  L(pAC(x,T(y))) -> pAC(L(pAC(x,y)),L(y))

Proof:
 AC-KBO Processor:
  precedence:
   T > L ~ one ~ pAC
   weight function:
    [L](x0) = 3x0 + 1,
    
    [T](x0) = 2x0 + 1,
    
    [one] = 8,
    
    [pAC](x0, x1) = x0 + x1 + 1
   problem:
    Equations:
     pAC(pAC(x2,x3),x4) -> pAC(x2,pAC(x3,x4))
     pAC(x2,x3) -> pAC(x3,x2)
     pAC(x2,pAC(x3,x4)) -> pAC(pAC(x2,x3),x4)
     pAC(x3,x2) -> pAC(x2,x3)
    TRS:
     
   Qed