Answer
34. Construct a truth table for each of these compound propositions
a) p $\oplus$ p
p
0 | 0
1 | 0
b) p $\oplus$ $\neg$p
p
0 | 1
1 | 1
c) p $\oplus$ $\neg$q
p , q
0 , 0 | 1
0 , 1 | 0
1 , 0 | 0
1 , 1 | 1
d) $\neg$p $\oplus$ $\neg$q
p , q
0 , 0 | 0
0 , 1 | 1
1 , 0 | 1
1 , 1 | 0
e) (p $\oplus$ q) $\vee$ (p $\oplus$ $\neg$q)
p , q
0 , 0 | 1
0 , 1 | 1
1 , 0 | 1
1 , 1 | 1
f) (p $\oplus$ q) $\land$ (p $\oplus$ $\neg$q)
p , q
0 , 0 | 0
0 , 1 | 0
1 , 0 | 0
1 , 1 | 0
Work Step by Step
It is important to first note that $\oplus$ denotes the operation "XOR", meaning Exclusive Or.
Exclusive Or is named as such because if either the first or the second proposition is true, but not both, then the resulting proposition is true.
In the truth tables that have been shown above, I use '0' for false, and '1' for true,
so it follows that:
0 $\oplus$ 0 = 0
0 $\oplus$ 1 = 1
1 $\oplus$ 0 = 1
1 $\oplus$ 1 = 0
These are all the possible combinations in which one could XOR two propositions, making it a truth table. A truth table simply shows all the possible permutation of variables in a compound proposition, and displays the result.
There are also the symbols $\vee$(OR), $\land$(AND), $\neg$(NOT)
AND only evaluates to true when both propositions are true, so
0 $\land$ 0 = 0
0 $\land$ 1 = 0
1 $\land$ 0 = 0
1 $\land$ 1 = 1
OR evaluates to true when either propositions are true, so
0 $\vee$ 0 = 0
0 $\vee$ 1 = 1
1 $\vee$ 0 = 1
1 $\vee$ 1 = 1
NOT inverts the proposition that it is given, so
$\neg$0 = 1
$\neg$1 = 0
Knowing all this, we can evaluate all of the possible permutations of each compound proposition with ease. Below is the evaluation of the possible compound propositions into true or false, which we then take the result of to put into our truth tables.
---a) p $\oplus$ p
p = 0:
0 $\oplus$ 0 = 0
p = 1:
1 $\oplus$ 1 = 0
b) p $\oplus$ $\neg$ p
p = 0:
0 $\oplus$ $\neg$0 = 0 $\oplus$ 1 = 1
p = 1:
1 $\oplus$ $\neg$1 = 1 $\oplus$ 0 = 1
c) p $\oplus$ $\neg$q
p = 0, q = 0:
0 $\oplus$ $\neg$0 = 0 $\oplus$ 1 = 1
p = 0, q = 1:
0 $\oplus$ $\neg$1 = 0 $\oplus$ 0 = 0
p = 1, q = 0:
1 $\oplus$ $\neg$0 = 1 $\oplus$ 1 = 0
p = 1, q = 1:
1 $\oplus$ $\neg$1 = 1 $\oplus$ 0 = 1
d) $\neg$p $\oplus$ $\neg$q
p = 0, q = 0
$\neg$0 $\oplus$ $\neg$0 = 1 $\oplus$ 1 = 0
p = 0, q = 1
$\neg$0 $\oplus$ $\neg$1 = 1 $\oplus$ 0 = 1
p = 1, q = 0
$\neg$1 $\oplus$ $\neg$0 = 0 $\oplus$ 1 = 1
p = 1, q = 1
$\neg$1 $\oplus$ $\neg$1 = 0 $\oplus$ 0 = 0
e) (p $\oplus$ q) $\vee$ (p $\oplus$ $\neg$q)
p = 0, q = 0
(0 $\oplus$ 0) $\vee$ (0 $\oplus$ $\neg$0) = 0 $\vee$ (0 $\oplus$ 1) = 0 $\vee$ 1 = 1
p = 0, q = 1
(0 $\oplus$ 1) $\vee$ (0 $\oplus$ $\neg$1) = 1 $\vee$ (0 $\oplus$ 0) = 1 $\vee$ 0 = 1
p = 1, q = 0
(1 $\oplus$ 0) $\vee$ (1 $\oplus$ $\neg$0) = 1 $\vee$ (1 $\oplus$ 1) = 1 $\vee$ 0 = 1
p = 1, q = 1
(1 $\oplus$ 1) $\vee$ (1 $\oplus$ $\neg$1) = 0 $\vee$ (1 $\oplus$ 0) = 0 $\vee$ 1 = 1
f) (p $\oplus$ q) $\land$ (p $\oplus$ $\neg$q)
p = 0, q = 0
(0 $\oplus$ 0) $\land$ (0 $\oplus$ $\neg$0) = 0 $\land$ (0 $\oplus$ 1) = 0 $\land$ 1 = 0
p = 0, q = 1
(0 $\oplus$ 1) $\land$ (0 $\oplus$ $\neg$1) = 1 $\land$ (0 $\oplus$ 0) = 1 $\land$ 0 = 0
p = 1, q = 0
(1 $\oplus$ 0) $\land$ (1 $\oplus$ $\neg$0) = 1 $\land$ (1 $\oplus$ 1) = 1 $\land$ 0 = 0
p = 1, q = 1
(1 $\oplus$ 1) $\land$ (1 $\oplus$ $\neg$1) = 0 $\land$ (1 $\oplus$ 0) = 0 $\land$ 1 = 0
