YES

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

Proof:
 DP Processor:
  DPs:
   c#(a(a(b(c(a(x1)))))) -> c#(x1)
   c#(a(a(b(c(a(x1)))))) -> c#(a(a(b(c(x1)))))
   c#(a(a(b(c(a(x1)))))) -> c#(c(a(a(b(c(x1))))))
  TRS:
   c(a(a(b(c(a(x1)))))) -> a(a(b(c(c(a(a(b(c(x1)))))))))
  Matrix Interpretation Processor: dim=5
   
   interpretation:
    [c#](x0) = [1 0 0 0 0]x0,
    
              [0 0 0 0 0]     [0]
              [0 0 0 0 0]     [0]
    [b](x0) = [0 0 1 0 0]x0 + [0]
              [0 1 0 0 0]     [0]
              [0 0 0 0 0]     [1],
    
              [0 0 1 0 1]  
              [0 0 0 1 0]  
    [c](x0) = [1 0 0 0 0]x0
              [0 0 0 0 1]  
              [0 1 1 0 1]  ,
    
              [1 0 0 1 0]     [0]
              [0 0 1 0 0]     [0]
    [a](x0) = [0 1 0 0 0]x0 + [0]
              [1 0 1 0 0]     [1]
              [1 1 0 0 0]     [0]
   orientation:
    c#(a(a(b(c(a(x1)))))) = [2 0 1 1 0]x1 + [2] >= [1 0 0 0 0]x1 = c#(x1)
    
    c#(a(a(b(c(a(x1)))))) = [2 0 1 1 0]x1 + [2] >= [1 0 0 1 0]x1 + [1] = c#(a(a(b(c(x1)))))
    
    c#(a(a(b(c(a(x1)))))) = [2 0 1 1 0]x1 + [2] >= [2 0 0 1 0]x1 = c#(c(a(a(b(c(x1))))))
    
                           [3 0 1 2 0]     [1]    [3 0 0 2 0]     [1]                                
                           [1 0 1 0 0]     [2]    [0 0 0 0 0]     [0]                                
    c(a(a(b(c(a(x1)))))) = [2 0 1 1 0]x1 + [2] >= [2 0 0 1 0]x1 + [0] = a(a(b(c(c(a(a(b(c(x1)))))))))
                           [2 0 1 1 0]     [1]    [1 0 0 1 0]     [1]                                
                           [3 0 1 2 0]     [1]    [3 0 0 2 0]     [0]                                
   problem:
    DPs:
     
    TRS:
     c(a(a(b(c(a(x1)))))) -> a(a(b(c(c(a(a(b(c(x1)))))))))
   Qed