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:
 Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {10,7,5,1}
   transitions:
    f0(4) -> 6*
    f0(3) -> 4*
    f0(2) -> 11*
    f0(8) -> 9*
    g1(15) -> 16*
    g1(26) -> 27*
    g1(30) -> 31*
    g1(28) -> 29*
    c0(2) -> 3*
    f2(43) -> 44*
    g0(11) -> 10*
    g0(4) -> 1*
    g0(6) -> 5*
    g0(2) -> 8*
    g0(9) -> 7*
    f30() -> 2*
    f1(13) -> 14*
    f1(27) -> 28*
    f1(34) -> 35*
    f1(14) -> 15*
    g2(42) -> 43*
    g2(44) -> 45*
    c1(12) -> 13*
    35 -> 28*
    7 -> 8*
    16 -> 1*
    45 -> 27*
    28 -> 42*
    11 -> 26*
    2 -> 12*
    6 -> 30*
    1 -> 8*
    31 -> 27*
    5 -> 10*
    10 -> 11*
    29 -> 27,34,10,11
  problem:
   
  Qed