YES
Time: 0.052

Problem:
 Equations:
  minAC(minAC(x3,x4),x5) -> minAC(x3,minAC(x4,x5))
  minAC(x3,x4) -> minAC(x4,x3)
  maxAC(maxAC(x3,x4),x5) -> maxAC(x3,maxAC(x4,x5))
  maxAC(x3,x4) -> maxAC(x4,x3)
  minAC(x3,minAC(x4,x5)) -> minAC(minAC(x3,x4),x5)
  minAC(x4,x3) -> minAC(x3,x4)
  maxAC(x3,maxAC(x4,x5)) -> maxAC(maxAC(x3,x4),x5)
  maxAC(x4,x3) -> maxAC(x3,x4)
 TRS:
  plus(x,0()) -> x
  plus(x,s(y)) -> s(plus(x,y))
  minAC(0(),y) -> 0()
  minAC(s(x),s(y)) -> s(minAC(x,y))
  maxAC(0(),y) -> y
  maxAC(s(x),s(y)) -> s(maxAC(x,y))

Proof:
 AC-KBO Processor:
  precedence:
   plus > s ~ 0 ~ maxAC ~ minAC
   weight function:
    [s](x0) = x0 + 10,
    
    [plus](x0, x1) = x0 + x1 + 4,
    
    [0] = 2,
    
    [maxAC](x0, x1) = x0 + x1 + 8,
    
    [minAC](x0, x1) = x0 + x1 + 12
   problem:
    Equations:
     minAC(minAC(x3,x4),x5) -> minAC(x3,minAC(x4,x5))
     minAC(x3,x4) -> minAC(x4,x3)
     maxAC(maxAC(x3,x4),x5) -> maxAC(x3,maxAC(x4,x5))
     maxAC(x3,x4) -> maxAC(x4,x3)
     minAC(x3,minAC(x4,x5)) -> minAC(minAC(x3,x4),x5)
     minAC(x4,x3) -> minAC(x3,x4)
     maxAC(x3,maxAC(x4,x5)) -> maxAC(maxAC(x3,x4),x5)
     maxAC(x4,x3) -> maxAC(x3,x4)
    TRS:
     
   Qed