Question

Which of the following proposition is a contradiction?
A. $\left( \sim p\vee \sim q \right)\vee \left( p\vee \sim q \right)$
B. $\left( p\to q \right)\vee \left( p\wedge \sim q \right)$
C. $\left( \sim p\wedge q \right)\wedge \left( \sim q \right)$
D. $\left( \sim p\wedge q \right)\vee \left( \sim q \right)$

Answer 1

Hint: We first define what a contradiction means for a proposition. Then we try to explain the signs used in the propositions. We then take Boolean logic or true-false logic to find out the outcomes of the propositions. If we get one true outcome then it can’t be a contradiction. The all false outcome will be considered as a contradiction.

Complete step-by-step answer:
The statement which one is false will be considered as a contradiction.
We have used three different notations. First, we define them and then try to describe the full statement.
Here $\sim $ defines the negation of a statement. $\to $ defines the implications of a statement. $\wedge $ defines the ‘and’, togetherness of events or statements. $\vee $ defines the ‘or’ of events or statements.
A proposition is a contradiction when all the outcomes possible are false or negative sense carriers.
That means if we take two events p and q as true-false then the outcome in our chosen proposition should always be false.
If one outcome of any proposition is true then it can’t be a contradiction.
In each case we are taking p and q as true-false events and the outcome as x.
So, in $x=\left( \sim p\vee \sim q \right)\vee \left( p\vee \sim q \right)$
When p false(F) and q false(F) we get x as $x=\left( T\vee T \right)\vee \left( F\vee T \right)=T\vee T=T$.
In $x=\left( p\to q \right)\vee \left( p\wedge \sim q \right)$
When p true(T) and q false(F) we get x as $x=\left( T\to F \right)\vee \left( T\wedge T \right)=F\vee T=T$.
$x=\left( \sim p\wedge q \right)\vee \left( \sim q \right)$
When p false(F) and q true(T) we get x as $x=\left( T\wedge T \right)\vee \left( F \right)=T\vee F=T$.
\[x=\left( \sim p\wedge q \right)\wedge \left( \sim q \right)\]
When p false(F) and q false(F) we get x as $x=\left( T\wedge F \right)\wedge \left( T \right)=F\wedge T=F$.
When p false(F) and q true(T) we get x as $x=\left( T\wedge T \right)\wedge \left( F \right)=T\wedge F=F$.
When p true(T) and q false(F) we get x as $x=\left( F\wedge F \right)\wedge \left( T \right)=F\wedge T=F$.
When p true(T) and q true(T) we get x as $x=\left( F\wedge T \right)\wedge \left( F \right)=F\wedge F=F$.
Therefore, the proposition \[x=\left( \sim p\wedge q \right)\wedge \left( \sim q \right)\] is a contradiction as its outcome is negative. The correct option is C.

So, the correct answer is “Option C”.

Note: We don’t need for the first three propositions to find out all their outcomes. We got one and that is enough to conclude that the proposition can’t be a contradiction. We need to have all the outcomes as false to find a contradiction. We can also work with Boolean 1 and 0 where we consider 1 as true and 0 as false.