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:
 DP Processor:
  DPs:
   g#(c(x1)) -> f#(c(x1))
   g#(c(x1)) -> g#(f(c(x1)))
   g#(f(c(x1))) -> f#(f(c(x1)))
   g#(f(c(x1))) -> g#(f(f(c(x1))))
   g#(g(x1)) -> f#(g(x1))
   g#(g(x1)) -> g#(f(g(x1)))
   f#(f(g(x1))) -> f#(x1)
   f#(f(g(x1))) -> g#(f(x1))
  TRS:
   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))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [f#](x0) = [3 0 0]x0,
    
    [g#](x0) = [2 2 3]x0,
    
              [0 0 2]  
    [f](x0) = [1 0 0]x0
              [1 0 0]  ,
    
              [0 0 0]     [0]
    [g](x0) = [1 1 1]x0 + [1]
              [1 1 1]     [1],
    
              [1 0 0]     [1]
    [c](x0) = [2 1 1]x0 + [2]
              [0 0 0]     [0]
   orientation:
    g#(c(x1)) = [6 2 2]x1 + [6] >= [3 0 0]x1 + [3] = f#(c(x1))
    
    g#(c(x1)) = [6 2 2]x1 + [6] >= [5 0 0]x1 + [5] = g#(f(c(x1)))
    
    g#(f(c(x1))) = [5 0 0]x1 + [5] >= [0] = f#(f(c(x1)))
    
    g#(f(c(x1))) = [5 0 0]x1 + [5] >= [4 0 0]x1 + [4] = g#(f(f(c(x1))))
    
    g#(g(x1)) = [5 5 5]x1 + [5] >= [0] = f#(g(x1))
    
    g#(g(x1)) = [5 5 5]x1 + [5] >= [4 4 4]x1 + [4] = g#(f(g(x1)))
    
    f#(f(g(x1))) = [6 6 6]x1 + [6] >= [3 0 0]x1 = f#(x1)
    
    f#(f(g(x1))) = [6 6 6]x1 + [6] >= [5 0 4]x1 = g#(f(x1))
    
               [0 0 0]     [0]    [0 0 0]     [0]              
    g(c(x1)) = [3 1 1]x1 + [4] >= [2 0 0]x1 + [3] = g(f(c(x1)))
               [3 1 1]     [4]    [2 0 0]     [3]              
    
                  [0 0 0]     [0]    [0 0 0]     [0]                 
    g(f(c(x1))) = [2 0 0]x1 + [3] >= [2 0 0]x1 + [3] = g(f(f(c(x1))))
                  [2 0 0]     [3]    [2 0 0]     [3]                 
    
               [0 0 0]     [0]    [0 0 0]     [0]              
    g(g(x1)) = [2 2 2]x1 + [3] >= [2 2 2]x1 + [3] = g(f(g(x1)))
               [2 2 2]     [3]    [2 2 2]     [3]              
    
                  [0 0 0]     [0]    [0 0 0]     [0]           
    f(f(g(x1))) = [2 2 2]x1 + [2] >= [2 0 2]x1 + [1] = g(f(x1))
                  [2 2 2]     [2]    [2 0 2]     [1]           
   problem:
    DPs:
     
    TRS:
     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))
   Qed