Discrete Structures

Lecture Notes, 13 Sep 2010

 Last Update: 13 September 2010 Note: or material is highlighted

1. Show that (p → (qr)) ≡ ((pq) → r):

To show that 2 propositions are logically equivalent,
we need to show that their truth tables have the same input-output (I-O) columns:

T-table for (p → (qr)):

p q r (qr) (p → (qr))
TTTTT
TTFFF
TFTTT
TFFTT
FTTTT
FTFFT
FFTTT
FFFTT

T-table for ((pq) → r):

p q r (pq) ((pq) → r)
TTTTT
TTFTF
TFTFT
TFFFT
FTTFT
FTFFT
FFTFT
FFFFT

Note that the I-O columns are the same, even though the intermediate columns are different.

2. One topic of CSE 191 is learning how to do proofs. Here's one:

Theorem:

Let A, B be propositions.
Then AB iff (AB) is a tautology.

I.e., two propositions (A, B) are equivalent iff their biconditional (AB) is a tautology.

Remarks:

1. Note that the book gives this as the definition of "logically equivalent".
This kind of thing happens frequently in math:
One mathematician's definition is another's theorem
(and, whenever that's the case, the first mathematician's theorem is the second mathematician's definition)!

2. Note that there are 3 different uses of a kind of "biconditional" in this theorem:

• The "≡" says that two propositions are logically equivalent.
• The "iff" says that two mathematical-English sentences
("AB" and "(AB) is a tautology") are necessary and sufficient for each other.
• The "↔" connects A with B to form a molecular proposition.

proof:

We have to show that a sentence of the form "P iff Q" is the case,
where P = "AB"
and where Q = "(AB) is a tautology".

To do that, we need to show two things:

1. If P, then Q
2. If Q, then P

1. To show: if P, then Q
we need to assume (or suppose, or make believe) that P is the case
and then show that—under that assumption (or under that supposition, or in that hypothetical situation)—Q is the case

Suppose AB, & show (AB) is a tautology:

Because AB, we can replace "≡" with its definition:
For all rows of the t-table, tval(A) = tval(B)

But, if we now add a column for (AB) to the t-table,
it will get filled in with Ts, because of the semantics of ↔

So, for all rows, tval(AB) = T.

But this is the definition of a tautology!

So, (AB) is a tautology
which is what we wanted to prove ("QED") in this first part of the proof

[This is where we ended today, but, for the sake of completeness, here's the rest of the proof, which we'll review in class on Wednesday.]

2. To show: if Q, then P
we need to assume that Q, and then show that P:

Suppose (AB) is a tautology, & show AB:

Because (AB) is a tautology, we can replace "is a tautology" by its definition:
for all rows, tval(AB) = T

But the only way for tval(AB) = T is for tval(A) = tval(B), by the semantics of ↔

So, for all rows, tval(A) = tval(B)

But this is the definition of logical equivalence!

So, AB,

which is QED for the second part of the proof.

Therefore, QED for the complete proof.
Next lecture…