YES

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

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
             [1 1]  
   [a](x0) = [1 1]x0,
   
             [1 1]     [1]
   [b](x0) = [0 0]x0 + [0],
   
             [1 0]  
   [c](x0) = [0 0]x0
  orientation:
                 [2 2]     [2]    [2 2]     [1]           
   a(a(b(x1))) = [2 2]x1 + [2] >= [0 0]x1 + [0] = b(a(x1))
   
              [1 1]     [1]    [1 1]     [1]              
   c(b(x1)) = [0 0]x1 + [0] >= [0 0]x1 + [0] = b(c(a(x1)))
   
           [1 1]      [1 1]                
   a(x1) = [1 1]x1 >= [1 1]x1 = a(c(a(x1)))
  problem:
   strict:
    c(b(x1)) -> b(c(a(x1)))
   weak:
    a(x1) -> a(c(a(x1)))
  Bounds Processor:
   bound: 1
   enrichment: match-rt
   automaton:
    final states: {4}
    transitions:
     b1(11) -> 12*
     a1(13) -> 14*
     a1(9) -> 10*
     c1(10) -> 11*
     c0(4) -> 4*
     b0(4) -> 4*
     a0(4) -> 4*
     12 -> 4*
     14 -> 4,10
     4 -> 9*
     11 -> 13*
   problem:
    strict:
     
    weak:
     a(x1) -> a(c(a(x1)))
   Qed