Discrete Structures
Lecture Notes Supplement, 22 Oct 2010
Last Update: 22 October 2010
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Here's the syntactic proof that:
[(A_{1} ∨ A_{2}) →
B] ≡ [(A_{1} → B)
∧ (A_{2} → B)]
To show that 2 propositions are logically equivalent,
we need to show that each implies the other.
 Show [(A_{1} ∨ A_{2})
→
B] → [(A_{1} → B)
∧ (A_{2} → B)]:
Suppose [(A_{1} ∨ A_{2})
→
B],
& show [(A_{1} → B)
∧ (A_{2} → B)]:
Show (a) (A_{1} → B)
and (b) (A_{2} → B):
 Suppose A_{1}, & show B:
From A_{1}, we can infer (A_{1} ∨
A_{2}), by Addition
∴ B, by MP (from our supposition that [(A_{1} ∨
A_{2})
→
B]).
 Suppose A_{2}, & show B:
Proof is similar, mutatis mutandis(*)
 Show [(A_{1} → B)
∧ (A_{2} → B)]
→ [(A_{1} ∨ A_{2}) →
B]:
Suppose [(A_{1} → B)
∧ (A_{2} → B)],
& show [(A_{1} ∨ A_{2}) →
B]:
Suppose (A_{1} ∨ A_{2}),
& show B:
 Suppose, by way of contradiction, that ¬B.
 We know (A_{1} → B), by Simplification from
one of our suppositions.
 Similarly, we also know (A_{2} → B).
 Now, it follows that ¬A_{1}, from (a),(b), by MT.
 ∴ A_{2}, by DS from another of our suppositions.
 ∴ B, by MP using (c)
 Contradiction! (We derived B from our assumption that ¬B.)
 ∴ B.
(I.e., our assumption that ¬B was wrong;
so we can conclude B.)
QED
Back to lecture of 10/22
(*)"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.
Text copyright © 2010 by William J. Rapaport
(rapaport@buffalo.edu)
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