Show that each of these conditional statements is a tautology using logical equivalences.a) (𝑝 ∧ 𝑞) → 𝑝b) ¬𝑝 → (𝑝 → 𝑞)c) (𝑝 ∧ 𝑞) → (𝑝 → 𝑞)

Sure, let's go through each of these conditional statements one by one:

a) (𝑝 ∧ 𝑞) → 𝑝

This statement can be rewritten using the logical equivalence of the conditional statement, which is ¬𝑝 ∨ 𝑞. So, (𝑝 ∧ 𝑞) → 𝑝 becomes ¬(𝑝 ∧ 𝑞) ∨ 𝑝.

Using De Morgan's law, ¬(𝑝 ∧ 𝑞) can be rewritten as ¬𝑝 ∨ ¬𝑞. So, ¬(𝑝 ∧ 𝑞) ∨ 𝑝 becomes (¬𝑝 ∨ ¬𝑞) ∨ 𝑝.

Using the associative law, we can rearrange this to ¬𝑝 ∨ (¬𝑞 ∨ 𝑝).

Finally, using the law of disjunction (which states that 𝑝 ∨ ¬𝑝 is always true), we can see that (¬𝑞 ∨ 𝑝) is always true, and therefore the entire statement is a tautology.

b) ¬𝑝 → (𝑝 → 𝑞)

This statement can be rewritten using the logical equivalence of the conditional statement, which is ¬𝑝 ∨ (𝑝 → 𝑞).

The inner conditional statement (𝑝 → 𝑞) can also be rewritten using the logical equivalence of the conditional statement, which gives us ¬𝑝 ∨ (¬𝑝 ∨ 𝑞).

Using the associative law, we can rearrange this to (¬𝑝 ∨ ¬𝑝) ∨ 𝑞.

Finally, using the law of idempotency (which states that 𝑝 ∨ 𝑝 is equivalent to 𝑝), we can simplify this to ¬𝑝 ∨ 𝑞.

Since ¬𝑝 ∨ 𝑞 is always true (because either 𝑝 is false, or 𝑝 is true and 𝑞 is true), the entire statement is a tautology.

c) (𝑝 ∧ 𝑞) → (𝑝 → 𝑞)

This statement can be rewritten using the logical equivalence of the conditional statement, which is ¬(𝑝 ∧ 𝑞) ∨ (𝑝 → 𝑞).

Using De Morgan's law, ¬(𝑝 ∧ 𝑞) can be rewritten as ¬𝑝 ∨ ¬𝑞. So, ¬(𝑝 ∧ 𝑞) ∨ (𝑝 → 𝑞) becomes (¬𝑝 ∨ ¬𝑞) ∨ (𝑝 → 𝑞).

The inner conditional statement (𝑝 → 𝑞) can also be rewritten using the logical equivalence of the conditional statement, which gives us (¬𝑝 ∨ ¬𝑞) ∨ (¬𝑝 ∨ 𝑞).

Using the associative law, we can rearrange this to ¬𝑝 ∨ (¬𝑞 ∨ (¬𝑝 ∨ 𝑞)).

Finally, using the law of disjunction (which states that 𝑝 ∨ ¬𝑝 is always true), we can see that (¬𝑞 ∨ (¬𝑝 ∨ 𝑞)) is always true, and therefore the entire statement is a tautology.