YES

Problem:
 g(c(x1)) -> g(f(c(x1)))
 g(f(c(x1))) -> g(f(f(c(x1))))
 g(g(x1)) -> g(f(g(x1)))
 f(f(g(x1))) -> g(f(x1))

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0  -&]  
   [f](x0) = [0  -&]x0,
   
             [0  0 ]  
   [g](x0) = [-& -&]x0,
   
             [0  -&]  
   [c](x0) = [1  0 ]x0
  orientation:
              [1  0 ]      [0  -&]                
   g(c(x1)) = [-& -&]x1 >= [-& -&]x1 = g(f(c(x1)))
   
                 [0  -&]      [0  -&]                   
   g(f(c(x1))) = [-& -&]x1 >= [-& -&]x1 = g(f(f(c(x1))))
   
              [0  0 ]      [0  0 ]                
   g(g(x1)) = [-& -&]x1 >= [-& -&]x1 = g(f(g(x1)))
   
                 [0 0]      [0  -&]             
   f(f(g(x1))) = [0 0]x1 >= [-& -&]x1 = g(f(x1))
  problem:
   g(f(c(x1))) -> g(f(f(c(x1))))
   g(g(x1)) -> g(f(g(x1)))
   f(f(g(x1))) -> g(f(x1))
  String Reversal Processor:
   c(f(g(x1))) -> c(f(f(g(x1))))
   g(g(x1)) -> g(f(g(x1)))
   g(f(f(x1))) -> f(g(x1))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 0]     [0]
     [f](x0) = [0 0 1]x0 + [1]
               [0 1 0]     [0],
     
               [1 0 1]  
     [g](x0) = [0 0 1]x0
               [0 0 1]  ,
     
               [1 0 0]  
     [c](x0) = [0 0 0]x0
               [0 1 0]  
    orientation:
                   [1 0 1]     [0]    [1 0 1]     [0]                 
     c(f(g(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = c(f(f(g(x1))))
                   [0 0 1]     [1]    [0 0 1]     [1]                 
     
                [1 0 2]      [1 0 2]                
     g(g(x1)) = [0 0 1]x1 >= [0 0 1]x1 = g(f(g(x1)))
                [0 0 1]      [0 0 1]                
     
                   [1 0 1]     [1]    [1 0 1]     [0]           
     g(f(f(x1))) = [0 0 1]x1 + [1] >= [0 0 1]x1 + [1] = f(g(x1))
                   [0 0 1]     [1]    [0 0 1]     [0]           
    problem:
     c(f(g(x1))) -> c(f(f(g(x1))))
     g(g(x1)) -> g(f(g(x1)))
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
                [0  0 ]  
      [f](x0) = [-& -&]x0,
      
                [0 1]  
      [g](x0) = [0 1]x0,
      
                [0  0 ]  
      [c](x0) = [-& 2 ]x0
     orientation:
                    [0  1 ]      [0  1 ]                   
      c(f(g(x1))) = [-& -&]x1 >= [-& -&]x1 = c(f(f(g(x1))))
      
                 [1 2]      [0 1]                
      g(g(x1)) = [1 2]x1 >= [0 1]x1 = g(f(g(x1)))
     problem:
      c(f(g(x1))) -> c(f(f(g(x1))))
     String Reversal Processor:
      g(f(c(x1))) -> g(f(f(c(x1))))
      Bounds Processor:
       bound: 1
       enrichment: match
       automaton:
        final states: {4}
        transitions:
         g1(13) -> 14*
         f1(12) -> 13*
         f1(11) -> 12*
         c1(10) -> 11*
         g0(4) -> 4*
         f0(4) -> 4*
         c0(4) -> 4*
         4 -> 10*
         14 -> 4*
       problem:
        
       Qed