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))
  Bounds Processor:
   bound: 2
   enrichment: match
   automaton:
    final states: {9,6,1}
    transitions:
     f30() -> 2*
     g0(10) -> 9*
     g0(5) -> 1*
     g0(2) -> 7*
     g0(8) -> 6*
     f0(7) -> 8*
     f0(2) -> 10*
     f0(4) -> 5*
     f0(3) -> 4*
     c0(2) -> 3*
     g1(15) -> 16*
     g1(17) -> 18*
     g1(19) -> 20*
     f1(16) -> 17*
     f1(23) -> 24*
     g2(31) -> 32*
     g2(33) -> 34*
     f2(32) -> 33*
     1 -> 7,9
     5 -> 19*
     6 -> 7*
     9 -> 10*
     10 -> 15*
     17 -> 31*
     18 -> 16,23,9,10
     20 -> 16*
     24 -> 17*
     34 -> 16*
   problem:
    
   Qed