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How To Prove It
1 Sentential Logic
1.3 Variables and Sets
1. Analyze the logical forms of the following statements:
(a) 3 is a common divisor of 6, 9, and 15. (Note: You did this in exercise 1 of Section 1.1, but you should be
able to give a better answer now.)
(Solution)
Let D (x,3) stand for “ x is divisible by 3.”
The entire statement would then be represented by the formula D(6,3)  D(9,3)  D(15,3).
(b) x is divisible by both 2 and 3 but not 4.
(Solution)
Let D ( x, y ) stand for “ x is divisible by y. ”
Thus “ x is divisible by2” would be represented by D (x,2).
Similarly, “ x is divisible by 3” by D (x,3), and“ x is divisible by 4” by D (x,4).
The entire statement would then be represented by the formula D( x,2)  D( x,3)  D( x,4).
(c) x and y are natural numbers, and exactly one of them is prime.
(Solution)
Let N (x) stand for “ x is a natural number.”
Let P (x ) stand for “ x is a prime number.”
Since exactly one of x and y is prime, x and y cannot be prime at the same time. Then it would lead us to the
following formula:
[ P( x)  P( y )]  [ P( x)  P( y )].
The entire statement would then be represented by the formula:
N ( x)  N ( y )  [{P( x)  P( y )}  {P( x)  P( y )}].
(d) x and y are men, and either x is taller than y or y is taller than x.
(Solution)
Let M (x) stand for “ x is a man.”
Let T ( x, y ) stand for “ x is taller than y. ”
Since either x is taller than y or y is taller than x, we would get the following formula:
T ( x, y )  T ( y, x).
The entire statement would then be represented by the formula: M ( x)  M ( y )  [T ( x, y )  T ( y, x)].
2. Write definitions using elementhood tests for the following sets:
(a) [Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto].
(Solution)
P  { p | p is a planet in the solar system } .
(b) [Brown, Columbia, Cornell, Dartmouth, Harvard, Princeton, University of Pennsylvania, Yale].
(Solution)
I  {i | i is an Ivy League school } .
(c) [1, 4, 9, 16, 25, 36, 49, ….].
(Solution)
S  {n 2 | n is a positive integer } .
3. Simplify the following statements. Which variables are free and which are bound? If the statement has no
free variables, say whether it is true or false.
(With MS Word Equation Editor, I was not able to type R to mean the set of real numbers.)
(a)  3  {x  R | 13  2 x  1}.
(Solution)
13  2 x  1
 2 x  12
x6
Therefore, the statement says  3  {x  R | x  6}.
Since  3 is a constant, this statement does not have a free variable, and x is the bound variable.
The statement is true.
(b) 5  {x  R | 13  2 x  c}.
(Solution)
13  2 x  c
 2 x  c  13
 c  13
x
2
 c  13 

Therefore, the statement says 5   x  R | x 
.
2 

Since 5 is a constant, this statement does not have a free variable, and x is the bound variable.
As we do not know the value of c, we cannot determine whether the statement is true with the given
information.
(c) w  {x  R | 13  2 x  c}.
(Solution)
From (b), we can conclude that x 
 c  13
.
2
 c  13
}.
2
In this statement, the free variable is w, and x is the bound variable.
Therefore, the statement says w  {x  R | x 
(d) 4  {x  { y | y is a prime number } | 13  2 x  1}. (It might make this statement easier to read if we let
P  { y | y is a prime number }; using this notation, we could rewrite the statement as
4  {x  P | 13  2 x  1}. )
(Solution)
Let P  { y | y is a prime number }.
From (a), we can conclude that x  6.
Therefore, the statement says 4  {x  P | x  6}, where P  { y | y is a prime number }.
Since 4 is a constant, the statement does not have a free variable, and x and y are the bound variables.
As 4 is not a prime, the statement is false.
4. What are the truth sets of the following statements? List a few elements of the truth set if you can.
(a) Elizabeth Taylor was once married to x.
(Solution)
{x | Elizabeth Taylor was married to x}  { Conrad Hilton Jr. (1950-1951), Michael Wilding (1952-1957),
Michael Todd (1957-1958), Eddie Fisher (1959-1964), Richard Burton (1964-1974) (1975-1976), John
Warner (1976-1982), and Larry Fortensky (1991-1996) } , from http://en.wikipedia.org/wiki/Elizabeth_Taylor,
as of writing of this (09/16/2007).
(b) x is a logical connective studied in Section 1.1.
(Solution)
{x | x is a logical connectives studied in Section 1.1 }  {,, and }.
(c) x is a real number and x 2  4 x  3  0.
(Solution)
x 2  4x  3  0
( x  1)( x  3)  0
 x  1or 3.
Therefore, {x | x  R, x 2  4 x  3  0}  {1,3}.
(d) x is a real number and x 2  2 x  3  0.
(Solution)
x 2  2x  3  0
 x  1  1  3  1  2i
Therefore, {x | x  R, x 2  2 x  3  0}  {}.
(e) 5  { y | x 2  y 2  50}.
(Solution)
x 2  y 2  50
x 2  50  y 2
Since 5  y, we can substitute y  5.
 x 2  50  5 2
x 2  50  25
x 2  25
( x  5)( x  5)  25
5 x  5
Therefore, the truth sets to the given statement is: {x | 5  x  5}.