YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 3
  interpretation:
             [0  0  -&]  
   [a](x0) = [0  0  -&]x0
             [1  1  0 ]  ,
   
             [0  -& -&]  
   [c](x0) = [-& -& 0 ]x0
             [-& 0  -&]  
  orientation:
                 [0  -& 0 ]             
   c(a(c(x1))) = [1  0  1 ]x1 >= x1 = x1
                 [0  -& 0 ]             
   
                 [1 1 0]      [0  0  -&]                   
   a(c(a(x1))) = [1 1 0]x1 >= [0  0  -&]x1 = a(a(a(a(x1))))
                 [2 2 1]      [1  1  0 ]                   
   
                                 
   x1 = x1 >= x1 = c(c(c(c(x1))))
                                 
  problem:
   strict:
    c(a(c(x1))) -> x1
   weak:
    x1 -> c(c(c(c(x1))))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [a](x0) = 8x0,
    
    [c](x0) = x0
   orientation:
    c(a(c(x1))) = 8x1 >= x1 = x1
    
    x1 = x1 >= x1 = c(c(c(c(x1))))
   problem:
    strict:
     
    weak:
     x1 -> c(c(c(c(x1))))
   Qed