YES
Time: 0.048

Problem:
 Equations:
  NAC(NAC(x3,x4),x5) -> NAC(x3,NAC(x4,x5))
  NAC(x3,x4) -> NAC(x4,x3)
  maxAC(maxAC(x3,x4),x5) -> maxAC(x3,maxAC(x4,x5))
  maxAC(x3,x4) -> maxAC(x4,x3)
  maxxAC(maxxAC(x3,x4),x5) -> maxxAC(x3,maxxAC(x4,x5))
  maxxAC(x3,x4) -> maxxAC(x4,x3)
  NAC(x3,NAC(x4,x5)) -> NAC(NAC(x3,x4),x5)
  NAC(x4,x3) -> NAC(x3,x4)
  maxAC(x3,maxAC(x4,x5)) -> maxAC(maxAC(x3,x4),x5)
  maxAC(x4,x3) -> maxAC(x3,x4)
  maxxAC(x3,maxxAC(x4,x5)) -> maxxAC(maxxAC(x3,x4),x5)
  maxxAC(x4,x3) -> maxxAC(x3,x4)
 TRS:
  h(L(x)) -> 0()
  h(NAC(x,y)) -> maxAC(h(x),h(y))
  maxAC(0(),x) -> x
  maxAC(x,0()) -> x
  maxAC(s(x),s(y)) -> s(maxxAC(x,y))
  maxxAC(0(),x) -> x
  maxxAC(x,0()) -> x
  maxxAC(s(x),s(y)) -> s(maxAC(x,y))

Proof:
 AC-KBO Processor:
  precedence:
   L > h > s ~ 0 ~ maxxAC ~ maxAC ~ NAC
   weight function:
    [s](x0) = 3x0 + 8,
    
    [0] = 15,
    
    [h](x0) = x0 + 2,
    
    [L](x0) = 2x0 + 13,
    
    [maxxAC](x0, x1) = x0 + x1 + 3,
    
    [maxAC](x0, x1) = x0 + x1 + 2,
    
    [NAC](x0, x1) = x0 + x1 + 5
   problem:
    Equations:
     NAC(NAC(x3,x4),x5) -> NAC(x3,NAC(x4,x5))
     NAC(x3,x4) -> NAC(x4,x3)
     maxAC(maxAC(x3,x4),x5) -> maxAC(x3,maxAC(x4,x5))
     maxAC(x3,x4) -> maxAC(x4,x3)
     maxxAC(maxxAC(x3,x4),x5) -> maxxAC(x3,maxxAC(x4,x5))
     maxxAC(x3,x4) -> maxxAC(x4,x3)
     NAC(x3,NAC(x4,x5)) -> NAC(NAC(x3,x4),x5)
     NAC(x4,x3) -> NAC(x3,x4)
     maxAC(x3,maxAC(x4,x5)) -> maxAC(maxAC(x3,x4),x5)
     maxAC(x4,x3) -> maxAC(x3,x4)
     maxxAC(x3,maxxAC(x4,x5)) -> maxxAC(maxxAC(x3,x4),x5)
     maxxAC(x4,x3) -> maxxAC(x3,x4)
    TRS:
     
   Qed