===================== Es. 2, 15/01/30 ===================

L ::= e | N L
N ::= a | b

f(e, B) = B e
f(N L, B) = N f(L, B)

g(e) = e
g(N L) = f(g(L), N)

Test:

    g(aabbe)
--> f(g(abbe), a)
--> f(f(g(bbe), a), a)
--> f(f(f(g(be), b), a), a)
--> f(f(f(f(g(e), b), b), a), a)))
--> f(f(f(f(e, b), b), a), a)))
--> f(f(f(be, b), a), a)))
--> f(f(bf(e, b), a), a)))
--> f(f(bbe, a), a)
--> f(bf(be, a), a)
--> f(bbf(e, a), a)
--> f(bbae, a)
--> bf(bae, a)
--> bbf(ae, a)
--> bbaf(e, a)
--> bbaae