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:
 Bounds Processor:
  bound: 0
  enrichment: match
  automaton:
   final states: {9,8,5,1}
   transitions:
    q0(7) -> 5*
    q0(4) -> 1*
    g0(6) -> 7*
    g0(2) -> 3*
    p0(4) -> 8*
    p0(7) -> 9*
    f40() -> 2*
    f0(2) -> 6*
    f0(3) -> 4*
  problem:
   
  Qed