YES

Problem:
 strict:
  a(b(a(x1))) -> a(b(b(a(x1))))
  b(a(b(x1))) -> b(a(a(b(x1))))
 weak:
  a(x1) -> a(a(a(x1)))
  b(x1) -> b(b(b(x1)))

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 2 0]     [0]
   [b](x0) = [0 0 0]x0 + [0]
             [0 1 0]     [1],
   
             [1 0 2]     [0]
   [a](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0]
  orientation:
                 [1 0 2]     [6]    [1 0 2]     [4]                 
   a(b(a(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(b(b(a(x1))))
                 [0 0 0]     [0]    [0 0 0]     [0]                 
   
                 [1 4 0]     [4]    [1 4 0]     [4]                 
   b(a(b(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(a(b(x1))))
                 [0 0 0]     [2]    [0 0 0]     [2]                 
   
           [1 0 2]     [0]    [1 0 2]     [0]              
   a(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(a(a(x1)))
           [0 0 0]     [0]    [0 0 0]     [0]              
   
           [1 2 0]     [0]    [1 2 0]     [0]              
   b(x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(b(b(x1)))
           [0 1 0]     [1]    [0 0 0]     [1]              
  problem:
   strict:
    b(a(b(x1))) -> b(a(a(b(x1))))
   weak:
    a(x1) -> a(a(a(x1)))
    b(x1) -> b(b(b(x1)))
  Bounds Processor:
   bound: 1
   enrichment: match-rt
   automaton:
    final states: {3}
    transitions:
     b0(3) -> 3*
     a0(3) -> 3*
     a1(11) -> 12*
     a1(14) -> 15*
     a1(26) -> 27*
     a1(24) -> 25*
     a1(25) -> 26*
     a1(10) -> 11*
     a1(20) -> 21*
     b1(9) -> 10*
     b1(12) -> 13*
     b1(28) -> 29*
     27 -> 25,3
     12 -> 20*
     21 -> 15*
     13 -> 10,14,3
     3 -> 24,9
     15 -> 11*
     10 -> 28*
     29 -> 13*
   problem:
    strict:
     
    weak:
     a(x1) -> a(a(a(x1)))
     b(x1) -> b(b(b(x1)))
   Qed