YES

Problem:
 a(b(x1)) -> b(a(x1))
 b(a(x1)) -> a(c(b(x1)))

Proof:
 DP Processor:
  DPs:
   a#(b(x1)) -> a#(x1)
   a#(b(x1)) -> b#(a(x1))
   b#(a(x1)) -> b#(x1)
   b#(a(x1)) -> a#(c(b(x1)))
  TRS:
   a(b(x1)) -> b(a(x1))
   b(a(x1)) -> a(c(b(x1)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [b#](x0) = [1 0 0]x0 + [1],
    
    [a#](x0) = [1 1 0]x0 + [1],
    
              [0 0 0]  
    [c](x0) = [0 0 1]x0
              [1 1 0]  ,
    
              [1 0 0]     [1]
    [a](x0) = [1 1 0]x0 + [1]
              [1 0 0]     [0],
    
              [1 0 0]     [1]
    [b](x0) = [0 1 0]x0 + [1]
              [1 0 0]     [0]
   orientation:
    a#(b(x1)) = [1 1 0]x1 + [3] >= [1 1 0]x1 + [1] = a#(x1)
    
    a#(b(x1)) = [1 1 0]x1 + [3] >= [1 0 0]x1 + [2] = b#(a(x1))
    
    b#(a(x1)) = [1 0 0]x1 + [2] >= [1 0 0]x1 + [1] = b#(x1)
    
    b#(a(x1)) = [1 0 0]x1 + [2] >= [1 0 0]x1 + [1] = a#(c(b(x1)))
    
               [1 0 0]     [2]    [1 0 0]     [2]           
    a(b(x1)) = [1 1 0]x1 + [3] >= [1 1 0]x1 + [2] = b(a(x1))
               [1 0 0]     [1]    [1 0 0]     [1]           
    
               [1 0 0]     [2]    [0 0 0]     [1]              
    b(a(x1)) = [1 1 0]x1 + [2] >= [1 0 0]x1 + [1] = a(c(b(x1)))
               [1 0 0]     [1]    [0 0 0]     [0]              
   problem:
    DPs:
     
    TRS:
     a(b(x1)) -> b(a(x1))
     b(a(x1)) -> a(c(b(x1)))
   Qed