Lecture Notes Supplement, 22 Oct 2010

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Here's the syntactic proof that:

[(A₁ ∨ A₂) → B] ≡ [(A₁ → B) ∧ (A₂ → B)]

To show that 2 propositions are logically equivalent,
we need to show that each implies the other.

1. Show [(A₁ ∨ A₂) → B] → [(A₁ → B) ∧ (A₂ → B)]:
   Suppose [(A₁ ∨ A₂) → B], & show [(A₁ → B) ∧ (A₂ → B)]:
   Show (a) (A₁ → B)
      and (b) (A₂ → B):
   1. Suppose A₁, & show B:
      From A₁, we can infer (A₁ ∨ A₂), by Addition
      ∴ B, by MP (from our supposition that [(A₁ ∨ A₂) → B]).

   2. Suppose A₂, & show B:
      Proof is similar, mutatis mutandis(*)

2. Show [(A₁ → B) ∧ (A₂ → B)] → [(A₁ ∨ A₂) → B]:
   Suppose [(A₁ → B) ∧ (A₂ → B)], & show [(A₁ ∨ A₂) → B]:
   Suppose (A₁ ∨ A₂), & show B:
   1. Suppose, by way of contradiction, that ¬B.
   2. We know (A₁ → B), by Simplification from one of our suppositions.
   3. Similarly, we also know (A₂ → B).
   4. Now, it follows that ¬A₁, from (a),(b), by MT.
   5. ∴ A₂, by DS from another of our suppositions.
   6. ∴ B, by MP using (c)
   7. Contradiction! (We derived B from our assumption that ¬B.)
   8. ∴ B.
      (I.e., our assumption that ¬B was wrong;
      so we can conclude B.)

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