`(p wedge q) to ((r wedgeq) wedge p)` is a

Options

A. Tautology
B. Contradiction
C. neither contradiction nor tautology
D. None of these

Answer

To determine the nature of the expression \((p \wedge q) \to ((r \wedge q) \wedge p)\), we will construct a truth table and analyze the results step by step.

### Step 1: Define the Variables
We have three variables: \(p\), \(q\), and \(r\). Each variable can be either true (T) or false (F).

### Step 2: Create the Truth Table
We will create a truth table with all combinations of truth values for \(p\), \(q\), and \(r\).

| \(p\) | \(q\) | \(r\) | \(p \wedge q\) | \(r \wedge q\) | \((r \wedge q) \wedge p\) | \((p \wedge q) \to ((r \wedge q) \wedge p)\) |
|-------|-------|-------|-----------------|----------------|---------------------------|----------------------------------------------|
| T | T | T | T | T | T | T |
| T | T | F | T | F | F | F |
| T | F | T | F | F | F | T |
| T | F | F | F | F | F | T |
| F | T | T | F | T | F | T |
| F | T | F | F | F | F | T |
| F | F | T | F | F | F | T |
| F | F | F | F | F | F | T |

### Step 3: Calculate \(p \wedge q\)
- \(p \wedge q\) is true only when both \(p\) and \(q\) are true.

### Step 4: Calculate \(r \wedge q\)
- \(r \wedge q\) is true only when both \(r\) and \(q\) are true.

### Step 5: Calculate \((r \wedge q) \wedge p\)
- This expression is true only when both \(r \wedge q\) and \(p\) are true.

### Step 6: Calculate the Implication \((p \wedge q) \to ((r \wedge q) \wedge p)\)
- The implication is true unless \(p \wedge q\) is true and \((r \wedge q) \wedge p\) is false.

### Step 7: Analyze the Results
From the truth table, we see that the final column \((p \wedge q) \to ((r \wedge q) \wedge p)\) has the following values:
- T, F, T, T, T, T, T, T

There is one instance where the implication is false (when \(p\) and \(q\) are both true but \(r\) is false). Since not all values are true and not all are false, the expression is neither a tautology nor a contradiction.

### Conclusion
The expression \((p \wedge q) \to ((r \wedge q) \wedge p)\) is neither a tautology nor a contradiction.