Show that `[(ptoq)^^(qto r)] to ( p to r)`is a tautology

Explanation

To show that the statement \([(p \to q) \land (q \to r)] \to (p \to r)\) is a tautology, we will construct a truth table.

### Step 1: Define the Variables
We have three variables: \(p\), \(q\), and \(r\). Each can be either true (T) or false (F). We will evaluate all possible combinations of truth values for these variables.

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

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

### Step 3: Calculate \(p \to q\), \(q \to r\), and \(p \to r\)
Next, we calculate the implications:

1. **\(p \to q\)**: This is false only when \(p\) is true and \(q\) is false.
2. **\(q \to r\)**: This is false only when \(q\) is true and \(r\) is false.
3. **\(p \to r\)**: This is false only when \(p\) is true and \(r\) is false.

Now, we can fill in the truth values for each implication:

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

### Step 4: Calculate \((p \to q) \land (q \to r)\)
Now we will calculate the conjunction of \(p \to q\) and \(q \to r\):

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

### Step 5: Calculate \([(p \to q) \land (q \to r)] \to (p \to r)\)
Finally, we will evaluate the main expression:

| \(p\) | \(q\) | \(r\) | \((p \to q) \land (q \to r)\) | \(p \to r\) | \([(p \to q) \land (q \to r)] \to (p \to r)\) |
|-------|-------|-------|--------------------------------|--------------|-----------------------------------------------|
| T | T | T | T | T | T |
| T | T | F | F | F | T |
| T | F | T | F | T | T |
| T | F | F | F | F | T |
| F | T | T | T | T | T |
| F | T | F | F | F | T |
| F | F | T | T | T | T |
| F | F | F | T | F | F |

### Step 6: Analyze the Results
The final column shows the results of \([(p \to q) \land (q \to r)] \to (p \to r)\). The values are all true except for the last case where \(p\), \(q\), and \(r\) are all false. However, since we are looking for a tautology, we need to ensure that the expression is true in all cases.

### Conclusion
Since the expression evaluates to true in all cases except one, we conclude that \([(p \to q) \land (q \to r)] \to (p \to r)\) is not a tautology.