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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?
in * Other Teaching Fields by (410 points)
You find your solution also in Aufgabe 4b of the exam from March 28, 2018 where the same problem was asked once more.

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Your solutions are shorter and can therefore be considered to be even better! Note, however, that it was not required to find the simplest solutions.
by (144k points)

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