Show that (PVQ)A] (1PA (QVR)) v ([PA]Q) v ([PA]R) is a tautology

To show that the expression:

\[
(P \lor Q) \land (\neg P \land (Q \lor R)) \lor (\neg P \land Q) \lor (\neg P \land R)
\]

is a tautology, we need to demonstrate that it is always true, regardless of the truth values of \(P\), \(Q\), and \(R\). A logical statement is a tautology if its truth table evaluates to true for all possible combinations of truth values of the variables involved.

We will construct the truth table step by step.

### 1. Breakdown of Components
- \( P \)
- \( Q \)
- \( R \)
- \( P \lor Q \)
- \( Q \lor R \)
- \( \neg P \)
- \( \neg P \land (Q \lor R) \)
- \( \neg P \land Q \)
- \( \neg P \land R \)

### 2. Construction of the Truth Table

| \( P \) | \( Q \) | \( R \) | \( P \lor Q \) | \( Q \lor R \) | \( \neg P \) | \( \neg P \land (Q \lor R) \) | \( \neg P \land Q \) | \( \neg P \land R \) | \( (P \lor Q) \land (\neg P \land (Q \lor R)) \lor (\neg P \land Q) \lor (\neg P \land R) \) |
|:------:|:------:|:------:|:----------:|:----------:|:--------:|:----------------------:|:-------------:|:-------------:|:----------------------------------------------------------------:|
|   T    |   T    |   T    |     T      |     T      |    F     |           F            |       F       |       F       |                                 F                                 |
|   T    |   T    |   F    |     T      |     T      |    F     |           F            |       F       |       F       |                                 F                                 |
|   T    |   F    |   T    |     T      |     T      |    F     |           F            |       F       |       F       |                                 F                                 |
|   T    |   F    |   F    |     T      |     F      |    F     |           F            |       F       |       F       |                                 F                                 |
|   F    |   T    |   T    |     T      |     T      |    T     |           T            |       T       |       F       |                                 T                                 |
|   F    |   T    |   F    |     T      |     T      |    T     |           T            |       T       |       F       |                                 T                                 |
|   F    |   F    |   T    |     F      |     T      |    T     |           T            |       F       |       T       |                                 T                                 |
|   F    |   F    |   F    |     F      |     F      |    T     |           F            |       F       |       F       |                                 F                                 |

### 3. Evaluation of the Final Expression

As we can see from the final column of the truth table, the expression evaluates to **True** in all cases where \(P\) is false. This confirms that the logical statement is **not** a tautology because there are combinations of truth values where the expression evaluates to false.



Show that (P ∨ Q) ∧ (¬P ∧ (Q ∨ R)) ∨ (¬P ∧ Q) ∨ (¬P ∧ R) is a tautology

Explore how to use truth tables to prove the logical expression (P ∨ Q) ∧ (¬P ∧ (Q ∨ R)) ∨ (¬P ∧ Q) ∨ (¬P ∧ R) is a tautology. This problem involves propositional logic and logical operators, suitable for advanced students in Grades 11-12.

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