YES
Time: 0.012

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-RPO Processor:
  precedence:
   plus > 0 > maxAC > minAC > s
  status:
   plus:mul  
  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