Tag: principle of mathematical induction

$(p\wedge ∼q)\wedge ∼p$

$(p\wedge ∼q)\vee  ∼p$

$(p\wedge ∼q)\vee ∼p$

none of these

If I do not have a Siberian Husky, then I do not have a dog.

If I have a dog, then I have a Siberian Husky.

If I do not have a dog, then I do not have a Siberian Husky.

If I do not have a Siberian Husky, then I have a dog.

$(p\vee q)$

$[p\wedge(q)]$

$p\wedge(q)$

$p\vee(q)$

a tautology

a contradiction

neither a tautology nor a contradiction

cannot come to any conclusion

you want of success and you find a way

you want of success and you do not find a way

if you do not want to succeed then you will find a way

if you want of success then you cannot find a way

$\left( { \sim p \vee q} \right) - \left( {p \vee \sim q} \right)$

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

$\left( {p \vee \sim q} \right) \wedge \left( {p \vee q} \right)$

$\left( { \sim p \vee \sim q} \right) \vee \left( {p \vee q} \right)$

Open the door

What an intelligent student!

Are you going to Delhi

All prime numbers are odd numbers

neither tautology nor contradiction

A tautology

A contradiction

Cannot be determined

$p \to q$

$p \to (q \vee p)$

$p \to (q \to p)$

$p \to (q \wedge p)$

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

$(p \rightarrow q) \equiv (~q \rightarrow ~p)$

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

$~(p \leftrightarrow q) \equiv (p \rightarrow q) \wedge (q \rightarrow p)$