Tautology and Contradiction

Which of the following Boolean expression is a tautology?

Let $p$ and $q$ be any two propositions.
Statement I: $(p\to q)\leftrightarrow q\vee ~p$ is a tautology.

Statement II: $~(~p\wedge q)\wedge (p\vee q)\leftrightarrow p$ is a fallacy.

The statement $[p\wedge (p\to q\left)\right]\to q$, is

The negation of $A\to \left(A\vee ~B\right)$ is

For two statements $\mathrm{p}$ and $\mathrm{q}$, prove the following relations of logical equivalence.

$\mathrm{p}\vee \mathrm{q}\to \mathrm{p}\equiv \mathrm{p}\vee ~\mathrm{q}$

For two statements $\mathrm{p}$ and $\mathrm{q}$, prove the following relations of logical equivalence.

$\left(\mathrm{p}\to \mathrm{q}\right)\to \mathrm{q}\equiv \mathrm{p}\vee \mathrm{q}$

$~\left(p\to q\right)\wedge \left(~p\wedge q\right)$ is

If $p$ and $q$ are two logical statements, then which of the following is not commutative ?

Let two statements $p$ and $q$ is defined as below

$p : 2\times 4=7$,

$q :$ Kota is in London.

Then truth values of these statements: $\left(i\right) p\leftrightarrow q$             $\left(ii\right) ~p\to q$              $\left(iii\right) \left(p\to q\right)\wedge q$ are

The statement $~\left(p\to q\right)\leftrightarrow \left(~p\vee ~q\right)$ is

The dual of $p\wedge ~p\equiv F$ is

The only statement among the following that is a tautology is

The logical equivalent statement of $p\leftrightarrow q$ is

$\left(p\Rightarrow q\right)\vee \left(q\Rightarrow r\right)$ is

$\sim \left(p\vee q\right)\vee \left(\sim p\wedge q\right)$ is equivalent to

The statement $~p\to \left(q\to p\right)$ is equivalent to

Let $p:$ Maths is interesting and $q:$ Maths is easy, then $p\Rightarrow \left(~p\vee q\right)$ is equivalent to

Let $p:$ Maths is interesting and $q:$ Maths is easy, then $p\Rightarrow \left(~p\vee q\right)$ is equivalent to

Without using truth table, prove that $\left[\left(p\vee q\right)\wedge ~p\right]\to q$ is a tautology.

Using the rules of logic, prove the following logical equivalences.

$~p\wedge q\equiv \left(p\vee q\right)\wedge ~p$