YES
Time: 0.012

Problem:
 Equations:
  meetAC(meetAC(x1,x2),x3) -> meetAC(x1,meetAC(x2,x3))
  meetAC(x1,x2) -> meetAC(x2,x1)
  meetAC(x1,meetAC(x2,x3)) -> meetAC(meetAC(x1,x2),x3)
  meetAC(x2,x1) -> meetAC(x1,x2)
 TRS:
  meetAC(x,x) -> x

Proof:
 Matrix Interpretation Processor:
  dimension: 2
  interpretation:
                                [1]
   [meetAC](x0, x1) = x0 + x1 + [0]
  orientation:
                 [2 0]    [1]         
   meetAC(x,x) = [0 2]x + [0] >= x = x
  problem:
   Equations:
    meetAC(meetAC(x1,x2),x3) -> meetAC(x1,meetAC(x2,x3))
    meetAC(x1,x2) -> meetAC(x2,x1)
    meetAC(x1,meetAC(x2,x3)) -> meetAC(meetAC(x1,x2),x3)
    meetAC(x2,x1) -> meetAC(x1,x2)
   TRS:
    
  Qed