YES

Problem:
 F(c(x)) -> G(x)
 G(x) -> F(x)
 c(x) -> x

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [G](x0) = 2x0,
   
   [F](x0) = x0,
   
   [c](x0) = 2x0
  orientation:
   F(c(x)) = 2x >= 2x = G(x)
   
   G(x) = 2x >= x = F(x)
   
   c(x) = 2x >= x = x
  problem:
   F(c(x)) -> G(x)
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  -&]  
    [G](x0) = [-& -&]x0,
    
              [3  -&]  
    [F](x0) = [-& 2 ]x0,
    
              [0 0]  
    [c](x0) = [0 1]x0
   orientation:
              [3 3]     [0  -&]        
    F(c(x)) = [2 3]x >= [-& -&]x = G(x)
   problem:
    
   Qed