Question is to show that we can modulate with the operator base {→, ¬} all the functions of the boolean operator base {¬, ∧, ∨}.

My solutions are a&b=!(a->!b), a|b=!a->b, but in the solutions they have way longer modulations for those two cases.

a ∨ b = (a → ¬b) → (¬b → a)

a ∧ b = ¬((¬a → b) → (b → ¬a))

Would my solution be sufficient or did i made a mistake?