To show that $(¬q ∧ (p ⇒ q)) ⇒ ¬q$ is a tautology, we need to prove that it is true in all possible cases using logical equivalences and properties. Let's break it down step by step, clearly labeling each step.

1. Express the implication using logical equivalence

Recall that an implication $p ⇒ q$ is equivalent to $(¬p ∨ q)$. So, we can rewrite the expression as follows:

$¬q ∧ (p ⇒ q) \equiv ¬q ∧ (¬p ∨ q)$

2. Distribute the conjunction

Now, distribute the conjunction $¬q ∧ (¬p ∨ q)$ using the distributive property:

$¬q ∧ (¬p ∨ q) \equiv (¬q ∧ ¬p) ∨ (¬q ∧ q)$

3. Simplify the conjunctions

Next, simplify the conjunctions:

• $¬q ∧ q$ is a contradiction, so it simplifies to false.
• $¬q ∧ ¬p$ remains as it is since there's no contradiction.

$(¬q ∧ ¬p) ∨ (¬q ∧ q) \equiv ¬q ∧ ¬p ∨ false$

4. Remove the false term

Since $A ∨ false \equiv A$, we can remove the false term:

$¬q ∧ ¬p ∨ false \equiv ¬q ∧ ¬p$

5. Express the implication

Now, we consider the whole expression $(¬q ∧ (p ⇒ q)) ⇒ ¬q$. Using the previous steps, we now have:

$(¬q ∧ ¬p) ⇒ ¬q$

An implication $A ⇒ B$ is false only when $A$ is true and $B$ is false. Let's analyze this:

• If $¬q ∧ ¬p$ is true, it means both $¬q$ and $¬p$ are true. But if $¬q$ is true, then the right-hand side $¬q$ is also true, making the implication true.
• If $¬q ∧ ¬p$ is false, the implication is vacuously true.

Conclusion:

In all cases, the implication $(¬q ∧ (p ⇒ q)) ⇒ ¬q$ is true, so it is a tautology.

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Do you want any further details or explanations? Here are some related questions you could explore:

1. What is a truth table, and how would it confirm this tautology?
2. How do logical equivalences simplify logical expressions?
3. What is the importance of the distributive property in logic?
4. Can all implications be converted using $p ⇒ q ≡ ¬p ∨ q$?
5. What are some examples of contradictions in logic?

Tip: Understanding common logical equivalences like De Morgan's laws and implication rules helps in solving complex logical expressions easily!