YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(a(a(x1)))) -> a#(b(a(x1)))
   a#(c(b(x1))) -> a#(x1)
   a#(c(b(x1))) -> a#(b(c(b(a(x1)))))
   a#(c(b(x1))) -> a#(a(b(c(b(a(x1))))))
  TRS:
   a(b(a(a(x1)))) -> c(b(a(b(a(x1)))))
   a(c(b(x1))) -> a(a(b(c(b(a(x1))))))
  Matrix Interpretation Processor: dim=4
   
   interpretation:
    [a#](x0) = [0 0 0 1]x0,
    
              [0 0 0 0]     [1]
              [0 0 0 0]     [0]
    [c](x0) = [0 0 0 0]x0 + [0]
              [0 1 0 0]     [1],
    
              [0 0 0 0]  
              [0 0 0 1]  
    [b](x0) = [0 0 1 0]x0
              [0 1 0 0]  ,
    
              [0 0 0 0]     [1]
              [1 0 1 0]     [0]
    [a](x0) = [1 0 1 0]x0 + [0]
              [0 0 1 0]     [0]
   orientation:
    a#(b(a(a(x1)))) = [1 0 1 0]x1 + [1] >= [1 0 1 0]x1 = a#(b(a(x1)))
    
    a#(c(b(x1))) = [0 0 0 1]x1 + [1] >= [0 0 0 1]x1 = a#(x1)
    
    a#(c(b(x1))) = [0 0 0 1]x1 + [1] >= [0] = a#(b(c(b(a(x1)))))
    
    a#(c(b(x1))) = [0 0 0 1]x1 + [1] >= [0] = a#(a(b(c(b(a(x1))))))
    
                     [0 0 0 0]     [1]    [0 0 0 0]     [1]                    
                     [1 0 1 0]     [1]    [0 0 0 0]     [0]                    
    a(b(a(a(x1)))) = [1 0 1 0]x1 + [1] >= [0 0 0 0]x1 + [0] = c(b(a(b(a(x1)))))
                     [1 0 1 0]     [1]    [1 0 1 0]     [1]                    
    
                  [1]    [1]                       
                  [1]    [1]                       
    a(c(b(x1))) = [1] >= [1] = a(a(b(c(b(a(x1))))))
                  [0]    [0]                       
   problem:
    DPs:
     
    TRS:
     a(b(a(a(x1)))) -> c(b(a(b(a(x1)))))
     a(c(b(x1))) -> a(a(b(c(b(a(x1))))))
   Qed