YES

Problem:
 p(f(f(x))) -> q(f(g(x)))
 p(g(g(x))) -> q(g(f(x)))
 q(f(f(x))) -> p(f(g(x)))
 q(g(g(x))) -> p(g(f(x)))

Proof:
 DP Processor:
  DPs:
   p#(f(f(x))) -> q#(f(g(x)))
   p#(g(g(x))) -> q#(g(f(x)))
   q#(f(f(x))) -> p#(f(g(x)))
   q#(g(g(x))) -> p#(g(f(x)))
  TRS:
   p(f(f(x))) -> q(f(g(x)))
   p(g(g(x))) -> q(g(f(x)))
   q(f(f(x))) -> p(f(g(x)))
   q(g(g(x))) -> p(g(f(x)))
  Usable Rule Processor:
   DPs:
    p#(f(f(x))) -> q#(f(g(x)))
    p#(g(g(x))) -> q#(g(f(x)))
    q#(f(f(x))) -> p#(f(g(x)))
    q#(g(g(x))) -> p#(g(f(x)))
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 2
    usable rules:
     
    interpretation:
     [q#](x0) = [0 0]x0 + [0],
     
     [p#](x0) = [0 0]x0 + [0],
     
               [2  0 ]     [2]
     [g](x0) = [-& 0 ]x0 + [1],
     
               [-& -&]     [1]
     [f](x0) = [-& 1 ]x0 + [2]
    orientation:
     p#(f(f(x))) = [-& 2 ]x + [3] >= [-& 1 ]x + [2] = q#(f(g(x)))
     
     p#(g(g(x))) = [4 2]x + [4] >= [-& 1 ]x + [3] = q#(g(f(x)))
     
     q#(f(f(x))) = [-& 2 ]x + [3] >= [-& 1 ]x + [2] = p#(f(g(x)))
     
     q#(g(g(x))) = [4 2]x + [4] >= [-& 1 ]x + [3] = p#(g(f(x)))
    problem:
     DPs:
      
     TRS:
      
    Qed