YES

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

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