The conditional statement `(p^^q) to p ` is

Options

A. a tautology
B. a fallacy
C. neither tautology nor fallacy
D. None of these

Explanation

To determine whether the conditional statement \((p \land q) \to p\) is a tautology, fallacy, or neither, we will analyze the logical expression step by step.

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

### Step 2: Determine the Possible Combinations
Since we have two variables, the total number of combinations of truth values is \(2^2 = 4\). The combinations are:
1. \(p = T\), \(q = T\)
2. \(p = T\), \(q = F\)
3. \(p = F\), \(q = T\)
4. \(p = F\), \(q = F\)

### Step 3: Create a Truth Table
We will create a truth table to evaluate the expression \((p \land q) \to p\).

| \(p\) | \(q\) | \(p \land q\) | \((p \land q) \to p\) |
|-------|-------|----------------|------------------------|
| T | T | T | T |
| T | F | F | T |
| F | T | F | T |
| F | F | F | T |

### Step 4: Evaluate \(p \land q\)
The conjunction \(p \land q\) is true only when both \(p\) and \(q\) are true:
- For \(p = T\), \(q = T\): \(T \land T = T\)
- For \(p = T\), \(q = F\): \(T \land F = F\)
- For \(p = F\), \(q = T\): \(F \land T = F\)
- For \(p = F\), \(q = F\): \(F \land F = F\)

### Step 5: Evaluate the Conditional Statement \((p \land q) \to p\)
The conditional statement \(A \to B\) is false only when \(A\) is true and \(B\) is false. In all other cases, it is true:
- For \(p = T\), \(q = T\): \(T \to T = T\)
- For \(p = T\), \(q = F\): \(F \to T = T\)
- For \(p = F\), \(q = T\): \(F \to F = T\)
- For \(p = F\), \(q = F\): \(F \to F = T\)

### Step 6: Conclusion
Since the result of \((p \land q) \to p\) is true for all combinations of truth values of \(p\) and \(q\), we conclude that the statement is a **tautology**.

### Final Answer
The conditional statement \((p \land q) \to p\) is a **tautology**.

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