Discrete Mathematics Assignment

<STUDENT NAME>
ASSIGNMENT
DISCRETE MATHEMATICS

1.
a.
The given symbol is the inversion of the OR logic and can be created by applying an inverter
on the OR truth table. The input signals are taken as p and q and the output is the last column.
p q p OR q p ↓q
0 0 0 1
0 1 1 0
1 0 1 0
1 1 1 0
b.
The given function is the combination of multiple NOR operations. It is developed step-by-
step in the table below.
p q p OR q p ↓q ( p ↓ q)↓( p ↓ q)
0 0 0 1 1
0 1 1 0 0
1 0 1 0 0
1 1 1 0 0
Observation: There is no change in the results. Whether we compute the function one time or
recursively, it does not change the results.
c.
We know that the NOT logic, i.e., negation has only one input. If we short both the inputs of
NOR logic, we result in the implementation of NEGATION.
p q f = p ↓ q
0 0 1
0 1 Invalid case (since both the
inputs can have only same
value)
1 0 Invalid case (since both the
inputs can have only same
value)
1 1 0
We can notice from the above truth table that the NOR logic behaves as the NOT logic.
If we apply the NOR logic on the inverted inputs of p and q, it behaves as the AND logic.

p q ~p ~q g= p ↓ q
0 0 1 1 0
0 1 1 0 0
1 0 0 1 0
1 1 0 0 1
From the above truth table, the AND logic is implemented using the NOR logic.
2.
P(x): x is connected to the network
Q(x): x has at least 100 terabytes of storage
a.
( P ( x ) ∧ Q ( x ) )
b.
P ( x ) ∧ Q ( x )
c.
P ( x ) ∨ Q ( x )
3.
P ( x , y ) : x2= y
a.
∃ x P(6 , x)
P ( 6 , x ) ⟹ 62=x
x=36
The truth table of this preposition will be a tautology.
b.
∃ x P ( x , 6 )
P ( x , 6 ) ⟹ x2=6
There exists no such whole number whose square is 6. Hence this statement is false.
c.
∀ x ∃ yP( x , y)
P ( x , y ) ⟹ x2= y
This statement is true as the square of a whole number is always a whole number.
d.
∃ y ∀ xP( x , y)