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I've tried to do this by assuming A, then using the law of excluded middle to get B ∨ ¬B. From here I plan on using or elimination to get A ∧ B then or introduction to get (A ∧ B) ∨ (A ∧ ¬B). I can get a proof of B → A ∧ B but not a proof of ¬B → A ∧ B. Is it possible to get a proof of ¬B → A ∧ B from what I have or do I need to prove A → (A ∧ B) ∨ (A ∧ ¬B) a different way? Thank you.

GXT
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    Use the distributivity of $\land$ over $\lor$. – Geoffrey Trang Feb 19 '20 at 16:24
  • It is quite sumple with LEM; use Disjunction Elim on $B \lor \lnot B$ to get $(A \land B)$ and $(A \land \lnot B)$ respectively. The result will follow using Disjunction Intro. – Mauro ALLEGRANZA Feb 19 '20 at 16:26
  • Thank you, I understand how to get B → A ∧ B but how can I get ¬B → A ∧ B so I can use Disjunction Elim on B∨¬B to get A∧B – GXT Feb 19 '20 at 16:52
  • From $B$ to to $B → A ∧ B$ and then $B → (A ∧ B) \lor (A ∧ ¬B)$ and from $¬B$ to to $B → A ∧ ¬B$ and then $¬B → (A ∧ ¬B) \lor (A ∧ B)$. After that, conclude by Disjunction Elim with $B ∨ ¬B$. – Mauro ALLEGRANZA Feb 20 '20 at 07:31

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