YES(?,O(n^1))

Problem:
 f(c(X,s(Y))) -> f(c(s(X),Y))
 g(c(s(X),Y)) -> f(c(X,s(Y)))

Proof:
 Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {4,3,2,1}
   transitions:
    f1(12) -> 4*
    f1(7) -> 1*
    c1(19,2) -> 12*
    c1(19,4) -> 12*
    c1(4,6) -> 12*
    c1(6,2) -> 7*
    c1(6,4) -> 7*
    c1(1,6) -> 12*
    c1(3,6) -> 12*
    c1(19,1) -> 12*
    c1(19,3) -> 12*
    c1(6,1) -> 7*
    c1(6,3) -> 7*
    c1(2,6) -> 12*
    s1(15) -> 19*
    s1(2) -> 6*
    s1(19) -> 19*
    s1(4) -> 6*
    s1(6) -> 6*
    s1(1) -> 6*
    s1(3) -> 6*
    f2(16) -> 4*
    c2(15,1) -> 16*
    c2(15,3) -> 16*
    c2(15,2) -> 16*
    c2(15,4) -> 16*
    c2(15,6) -> 16*
    f0(2) -> 1*
    f0(4) -> 1*
    f0(1) -> 1*
    f0(3) -> 1*
    s2(15) -> 15*
    s2(2) -> 15*
    s2(4) -> 15*
    s2(1) -> 15*
    s2(3) -> 15*
    c0(3,1) -> 2*
    c0(3,3) -> 2*
    c0(4,2) -> 2*
    c0(4,4) -> 2*
    c0(1,2) -> 2*
    c0(1,4) -> 2*
    c0(2,1) -> 2*
    c0(2,3) -> 2*
    c0(3,2) -> 2*
    c0(3,4) -> 2*
    c0(4,1) -> 2*
    c0(4,3) -> 2*
    c0(1,1) -> 2*
    c0(1,3) -> 2*
    c0(2,2) -> 2*
    c0(2,4) -> 2*
    s0(2) -> 3*
    s0(4) -> 3*
    s0(1) -> 3*
    s0(3) -> 3*
    g0(2) -> 4*
    g0(4) -> 4*
    g0(1) -> 4*
    g0(3) -> 4*
  problem:
   
  Qed