Exercice 1.1:

E1 = {f(Y, g(V), g(a) + V); f(Z, g(g(b)), X + g(b))}, X, Y, V variables.
V -> g(b)
X -> g(a)
Y -> Z

E2 = {f(a, g(V), X + V); f(a, g(g(V)), g(a) + g(b))}, X, V variables.
Pas possible:
V -> g(V)

E3 = {f([X|[X|R]]); f([a, a, b])} X, R variables
X -> a
R -> [b]

E4 = {couleur(X); couleur([Z|[a|T]])}, X, Z, T variables.
X -> [Z|[a|T]]

E5 = {f(g(k(b)), y), f(y, g(x))}, x, y variables.
y -> g(k(b))
y -> g(x)
x -> k(b)

E6 = {(f(y, k(y), g(x)), f(k(x), k(y), y))}, x, y variables.
Pas possible:
y -> k(x)
y -> g(x)

E7 = {f(x, y) = f(u, v); x = g(y); y = g(z); u = g(t); t = g(w)}, pas de constantes, que des variables.
x = u = g(t) = g(y)
y = v = g(z)
t = g(w)

E8 = {(f(g(x), x), f(y, g(z))); (g(x), y)}, x, y, z variables.
Pas possible
f(g(x), x) = g(x)