Evaluating [¬(p → Q) ˅ (p ↔ Q)] ˄ [(¬p → Q) ˅ ¬p] Formula Tautology Or Contradiction
Hey guys! Today, we're diving into the fascinating world of propositional logic. We've got a formula here: [¬(p → q) ˅ (p ↔ q)] ˄ [(¬p → q) ˅ ¬p], and our mission, should we choose to accept it (and we do!), is to evaluate it using the shorthand method, or truth tables, and determine whether it's a tautology (always true), a contradiction (always false), or neither. Buckle up, because we're about to break this down step-by-step!
1. Understanding the Basics
Before we jump into the nitty-gritty, let's quickly recap some fundamental concepts. In propositional logic, we deal with statements that can be either true or false. These statements are represented by variables, often p, q, r, and so on. We then combine these statements using logical connectives to form more complex formulas.
The key connectives we'll be using today are:
- ¬ (Negation): This simply reverses the truth value. If p is true, then ¬p is false, and vice-versa.
- → (Conditional): This represents "if...then". p → q is only false when p is true and q is false. Think of it as a promise; the promise is broken only if you fulfill the condition (p is true) but don't deliver the result (q is false).
- ↔ (Biconditional): This means "if and only if". p ↔ q is true when p and q have the same truth value (both true or both false).
- ˅ (Disjunction): This means "or". p ˅ q is true if either p is true, q is true, or both are true. It's only false if both p and q are false.
- ˄ (Conjunction): This means "and". p ˄ q is true only if both p and q are true.
With these definitions in our toolbelt, we're ready to tackle our formula!
2. Constructing the Truth Table
The truth table is our secret weapon for evaluating logical formulas. It systematically lists all possible combinations of truth values for our variables and then calculates the truth value of the entire formula for each combination. Since we have two variables, p and q, we'll have 2^2 = 4 rows in our truth table.
Here's the basic structure of our truth table:
| p | q | ¬(p → q) | (p ↔ q) | ¬(p → q) ˅ (p ↔ q) | ¬p | (¬p → q) | ¬p → q) ˅ ¬p | [¬(p → q) ˅ (p ↔ q)] ˄ [(¬p → q) ˅ ¬p] |
|---|---|---|---|---|---|---|---|---|
| T | T | |||||||
| T | F | |||||||
| F | T | |||||||
| F | F |
Now, let's fill in the table step-by-step.
Step 2.1: Filling in the Basic Truth Values
The first two columns, p and q, simply list all possible combinations of true (T) and false (F):
| p | q |
|---|---|
| T | T |
| T | F |
| F | T |
| F | F |
Step 2.2: Evaluating (p → q)
Remember, p → q is only false when p is true and q is false. So, let's fill in the column for (p → q):
| p | q | (p → q) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
Step 2.3: Evaluating ¬(p → q)
Now we negate the values we just calculated for (p → q). If (p → q) is true, ¬(p → q) is false, and vice-versa:
| p | q | (p → q) | ¬(p → q) |
|---|---|---|---|
| T | T | T | F |
| T | F | F | T |
| F | T | T | F |
| F | F | T | F |
Step 2.4: Evaluating (p ↔ q)
Remember, p ↔ q is true when p and q have the same truth value. Let's fill in the column:
| p | q | (p ↔ q) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | T |
Step 2.5: Evaluating [¬(p → q) ˅ (p ↔ q)]
This is the first main part of our formula. We're taking the disjunction (OR) of ¬(p → q) and (p ↔ q). Remember, the disjunction is true if either or both of the operands are true:
| p | q | ¬(p → q) | (p ↔ q) | ¬(p → q) ˅ (p ↔ q) |
|---|---|---|---|---|
| T | T | F | T | T |
| T | F | T | F | T |
| F | T | F | F | F |
| F | F | F | T | T |
Step 2.6: Evaluating ¬p
This is a straightforward negation of p:
| p | q | ¬p |
|---|---|---|
| T | T | F |
| T | F | F |
| F | T | T |
| F | F | T |
Step 2.7: Evaluating (¬p → q)
This is a conditional statement where the antecedent is ¬p and the consequent is q. Remember, it's only false when ¬p is true and q is false:
| p | q | ¬p | (¬p → q) |
|---|---|---|---|
| T | T | F | T |
| T | F | F | T |
| F | T | T | T |
| F | F | T | F |
Step 2.8: Evaluating [(¬p → q) ˅ ¬p]
Now we take the disjunction (OR) of (¬p → q) and ¬p:
| p | q | ¬p | (¬p → q) | (¬p → q) ˅ ¬p |
|---|---|---|---|---|
| T | T | F | T | T |
| T | F | F | T | T |
| F | T | T | T | T |
| F | F | T | F | T |
Step 2.9: Evaluating the Entire Formula [¬(p → q) ˅ (p ↔ q)] ˄ [(¬p → q) ˅ ¬p]
Finally, we take the conjunction (AND) of the two main parts we've calculated: [¬(p → q) ˅ (p ↔ q)] and [(¬p → q) ˅ ¬p]. Remember, the conjunction is only true if both operands are true:
| p | q | ¬(p → q) ˅ (p ↔ q) | (¬p → q) ˅ ¬p | [¬(p → q) ˅ (p ↔ q)] ˄ [(¬p → q) ˅ ¬p] |
|---|---|---|---|---|
| T | T | T | T | T |
| T | F | T | T | T |
| F | T | F | T | F |
| F | F | T | T | T |
3. Determining Tautology, Contradiction, or Neither
Now for the grand finale! We look at the final column of our truth table, which represents the truth value of the entire formula for each combination of p and q.
- If all the values in the final column are true (T), the formula is a tautology.
- If all the values in the final column are false (F), the formula is a contradiction.
- If the final column contains a mix of true and false values, the formula is neither a tautology nor a contradiction.
Looking at our final column:
| [¬(p → q) ˅ (p ↔ q)] ˄ [(¬p → q) ˅ ¬p] |
|---|
| T |
| T |
| F |
| T |
We see a mix of true and false values. Therefore, the formula [¬(p → q) ˅ (p ↔ q)] ˄ [(¬p → q) ˅ ¬p] is neither a tautology nor a contradiction.
4. Conclusion
And there you have it! We've successfully evaluated the formula [¬(p → q) ˅ (p ↔ q)] ˄ [(¬p → q) ˅ ¬p] using a truth table and determined that it is neither a tautology nor a contradiction. This step-by-step approach can be applied to any logical formula, allowing you to break down complex expressions into manageable chunks. Remember, practice makes perfect, so keep those truth tables coming!
Key Takeaways:
- Truth tables are a powerful tool for evaluating logical formulas.
- Understanding the definitions of logical connectives is crucial.
- A tautology is always true, a contradiction is always false, and a formula can be neither.
I hope this helped you guys! Let me know if you have any more logic puzzles you want to solve together!