YES
Time: 0.132

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:
 Matrix Interpretation Processor:
  dimension: 1
  interpretation:
   [s](x0) = x0 + 2,
   
   [plus](x0, x1) = x0 + 3x1 + 1,
   
   [0] = 0,
   
   [maxAC](x0, x1) = x0 + x1 + 1,
   
   [minAC](x0, x1) = x0 + x1 + 8
  orientation:
   plus(x,0()) = x + 1 >= x = x
   
   plus(x,s(y)) = x + 3y + 7 >= x + 3y + 3 = s(plus(x,y))
   
   minAC(0(),y) = y + 8 >= 0 = 0()
   
   minAC(s(x),s(y)) = x + y + 12 >= x + y + 10 = s(minAC(x,y))
   
   maxAC(0(),y) = y + 1 >= y = y
   
   maxAC(s(x),s(y)) = x + y + 5 >= x + y + 3 = s(maxAC(x,y))
  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