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I am stuck on one part of this question.

Suppose we have the single binary connective →, plus the symbol for absurdity ⊥. Using just these expressions, see if you can find a way to express ¬P, P∧Q, and P∨Q.

I know that for ¬P the answer is P→⊥

I know that for P∨Q the answer is (P→⊥)→Q

I am stuck on P∧Q..

And help would be great. Thanks!

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    Since φ∧ψ is equivalent with ¬(¬φv¬ψ), and since you know how to do ¬φ and φvψ, you should be able to do P∧Q as well. – MarkOxford Aug 3 '17 at 19:21
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You can combine the expressions for ¬ and v, as suggested by MarkOxford, but that yields a rather complicated expression. You get an easier solution if you start by developing an expression for ¬(P ∧ Q), namely

P → (Q → ⊥)

By negating this, we get an expression for P ∧ Q, namely

(P → (Q → ⊥)) → ⊥

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