Discrete Structures

# Lecture Notes Supplement, 22 Oct 2010

 Last Update: 22 October 2010 Note: or material is highlighted

Here's the syntactic proof that:

[(A1A2) → B] ≡ [(A1B) ∧ (A2B)]

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

1. Show [(A1A2) → B] → [(A1B) ∧ (A2B)]:
Suppose [(A1A2) → B], & show [(A1B) ∧ (A2B)]:
Show (a) (A1B)
and (b) (A2B):

1. Suppose A1, & show B:
From A1, we can infer (A1A2), by Addition
B, by MP (from our supposition that [(A1A2) → B]).

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

2. Show [(A1B) ∧ (A2B)] → [(A1A2) → B]:
Suppose [(A1B) ∧ (A2B)], & show [(A1A2) → B]:
Suppose (A1A2), & show B:
1. Suppose, by way of contradiction, that ¬B.
2. We know (A1B), by Simplification from one of our suppositions.
3. Similarly, we also know (A2B).
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.)
QED

(*)"mutatis mutandis" is Latin for: "changing what needs to be changed".
As used here, it's mathematical jargon meaning: Repeat the same proof as before, but change whatever needs to be changed, e.g., subscripts, etc.