YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [b](x0) = x0 + 2,
   
   [a](x0) = x0 + 1,
   
   [c](x0) = 2x0
  orientation:
   a(a(a(a(x1)))) = x1 + 4 >= x1 + 4 = b(b(x1))
   
   b(b(a(a(x1)))) = x1 + 6 >= x1 + 6 = a(a(b(b(x1))))
   
   b(b(b(b(c(c(x1)))))) = 4x1 + 8 >= 4x1 + 8 = c(c(a(a(x1))))
   
   b(b(b(b(x1)))) = x1 + 8 >= x1 + 6 = a(a(a(a(a(a(x1))))))
   
   c(c(a(a(x1)))) = 4x1 + 8 >= 4x1 + 6 = b(b(a(a(c(c(x1))))))
  problem:
   a(a(a(a(x1)))) -> b(b(x1))
   b(b(a(a(x1)))) -> a(a(b(b(x1))))
   b(b(b(b(c(c(x1)))))) -> c(c(a(a(x1))))
  KBO Processor:
   weight function:
    w0 = 1
    w(b) = w(a) = 1
    w(c) = 0
   precedence:
    c > b > a
   problem:
    
   Qed