Logical equivalence - class-XII

Evaluating the expression using truth tables or logical equivalences shows that it evaluates to false for all truth values of the variables. The components simplify such that a proposition and its negation are strictly contradicted, yielding a contradiction.

$\because [(p \wedge p ) \rightarrow q ] \rightarrow p \equiv ( p

\rightarrow q ) \rightarrow p   (\because p \wedge p \equiv p)$

$\because (\sim p  \vee \sim q) \vee (p  \vee \sim q) $

Since, all the entries in the last column has true value. So, the given statement is a tautology

So the result of the Truth table show that $p\Rightarrow \sim p$ is neither a tautology not a contradiction

Truth table is,

Hence given statement is contradiction.

The statement (p -> q) -> ((~p -> q) -> q) is a tautology. It can be verified using a truth table or logical equivalences.

Contradiction is a preposition which is always false(F).
Here $\sim \equiv negation$ and $\wedge \equiv AND$
Let $ x=(\sim p \wedge q) \wedge (\sim q)$
If $p=F$ and $q=F$ then $x=F$
If $p=F$ and $q=T$ then $x=F$
If $p=T$ and $q=F$ then $x=F$
If $p=T$ and $q=T$ then $x=F$
Hence option (c) is correct

The expression (p -> q) ^ (q -> r) -> (p -> r) is the Law of Hypothetical Syllogism, which is a fundamental tautology in propositional logic.

The statement

This is identical to the previous question; the expression (p -> q) ^ (q -> r) -> (p -> r) is a tautology.

A tautology is a compound statement that is always true regardless of the truth values of its components. For option C, the disjunction of any proposition p with a true statement t always results in true, satisfying the definition of a tautology.

p v (~p v q) is equivalent to (p v ~p) v q, which is t v q, which is always true (t). Thus, it is a tautology.

Option B represents modus ponens, which is a universally valid argument form and thus a tautology. By constructing a truth table, we find that the conditional statement always evaluates to true.

Using distributive and absorption laws, simplify the expression step by step. Distributing the terms or factoring out common components yields the disjunction of p and q.

Analyzing the expression reveals an internal contradiction because it forces both p and not p to be true simultaneously. Therefore, the entire conjunction is always false, making it a contradiction.

Option C represents the standard logical rule of detachment, or modus ponens, which is a tautology. Evaluating its truth table yields true in every row.

$P:{ 2 }^{ 2 }=5$

The expression (p ^ q) v q is equivalent to (p v q) v q. By associativity, this is ~p v (q v q), which is ~p v T, which is T. Thus, it is a tautology.

Statement I is true because the expression contains contradictory sub-parts making it a fallacy or contradiction. Statement II is true because a conditional is logically equivalent to its contrapositive, making it a tautology; however, Statement II does not explain Statement I.

The expression contains a conjunction of a statement and its direct negation through De Morgan expansion of the second term, ensuring it is always false. Thus, the statement is a contradiction.

(p ^ q) -> p is a tautology because if (p ^ q) is true, then p must be true. If (p ^ q) is false, the implication is vacuously true.

When $p$ and $q$ both are true then 

A contradiction is a statement that is always false. Option C, (p ^ q) ^ (q), simplifies to ~p ^ (q ^ ~q), which is ~p ^ F, which is always false.

A.   $p \wedge (\sim c )\equiv p \wedge t  \equiv p$

Truth table,

Hence option 'C' is the correct choice.

If $p$ is a true statement, then $p \rightarrow p$ is true.

We check for contradiction for all the given options.

All F so this is a contradiction.

When $p$ and $q$ both are true then

A tautology is a statement that is always true.
The statement $ p\Longrightarrow p\vee q $ is read as " if p is true, then either p or q is true. " as the symbol $ \vee  $ denotes OR.
Hence, the given statement is a tautology.

A,C,D obvious and for B
p $\rightarrow$ q is same as $\sim$ q $\rightarrow$ $\sim$p
$\therefore$ it is tautology not contradiction.

Tautology is the preposition which is always true and contradiction is a preposition which is always false.
Here $\wedge \equiv AND, \vee \equiv OR$, $T$=true and $F$=false
Given statement=$p$, tautology=$t$ and contradiction=$c$
a)$p\wedge t\equiv p$ irrespective of value of $p$ as $t=T$ always
And if $p=T$ then $T\wedge T=T$ and if $p=F$ then $F\wedge T=F$
Thus option (a) is correct.
b)$p\wedge c\equiv c$ irrespective of value of $p$ as$ c=F$ always
And if $p=T$ then $T\wedge F=F$ and if $p=F$ then $F\wedge F=F$
Thus option (b) is correct
c)$p\vee t\equiv c$
If $p=T$ then $T\vee T\equiv T\neq c$ and if $p=F$ then $F\vee T\equiv T \neq c$
Thus option (c) is not correct
d)$p\vee c\equiv p$
If $p=T$ then $T\vee F\equiv T$ and if $p=F$ then $F\vee F\equiv F$
Thus it depends on the value of $p$
Hence option (d) is correct

$\left( p\vee q \right) \rightarrow q$