YES
Time: 0.154

Problem:
 Equations:
  fAC(fAC(x3,x4),x5) -> fAC(x3,fAC(x4,x5))
  fAC(x3,x4) -> fAC(x4,x3)
  fAC(x3,fAC(x4,x5)) -> fAC(fAC(x3,x4),x5)
  fAC(x4,x3) -> fAC(x3,x4)
 TRS:
  fAC(fAC(i(x),i(y1)),y2) -> fAC(i(fAC(x,y1)),y2)
  fAC(one(),x) -> x
  phi(one(),x) -> x
  fAC(i(x),x) -> one()
  i(one()) -> one()
  i(i(x)) -> x
  phi(x,phi(y1,y2)) -> phi(fAC(x,y1),y2)
  fAC(i(fAC(x,y1)),y1) -> i(x)
  i(fAC(i(y1),x)) -> fAC(i(x),y1)
  fAC(i(x),i(y1)) -> i(fAC(x,y1))

Proof:
 Matrix Interpretation Processor:
  dimension: 1
  interpretation:
   [phi](x0, x1) = 2x0 + 4x1 + 4,
   
   [one] = 0,
   
   [i](x0) = 2x0 + 8,
   
   [fAC](x0, x1) = x0 + x1 + 1
  orientation:
   fAC(fAC(i(x),i(y1)),y2) = 2x + 2y1 + y2 + 18 >= 2x + 2y1 + y2 + 11 = fAC(i(fAC(x,y1)),y2)
   
   fAC(one(),x) = x + 1 >= x = x
   
   phi(one(),x) = 4x + 4 >= x = x
   
   fAC(i(x),x) = 3x + 9 >= 0 = one()
   
   i(one()) = 8 >= 0 = one()
   
   i(i(x)) = 4x + 24 >= x = x
   
   phi(x,phi(y1,y2)) = 2x + 8y1 + 16y2 + 20 >= 2x + 2y1 + 4y2 + 6 = phi(fAC(x,y1),y2)
   
   fAC(i(fAC(x,y1)),y1) = 2x + 3y1 + 11 >= 2x + 8 = i(x)
   
   i(fAC(i(y1),x)) = 2x + 4y1 + 26 >= 2x + y1 + 9 = fAC(i(x),y1)
   
   fAC(i(x),i(y1)) = 2x + 2y1 + 17 >= 2x + 2y1 + 10 = i(fAC(x,y1))
  problem:
   Equations:
    fAC(fAC(x3,x4),x5) -> fAC(x3,fAC(x4,x5))
    fAC(x3,x4) -> fAC(x4,x3)
    fAC(x3,fAC(x4,x5)) -> fAC(fAC(x3,x4),x5)
    fAC(x4,x3) -> fAC(x3,x4)
   TRS:
    
  Qed