To solve the problem, we need to determine the mathematical operators \( o+ \) and \( ox \) such that the expression \( p \, o+ \, (p \, ox \, q) \) is not a tautology. A tautology is a statement that is always true, regardless of the truth values of its components. Therefore, we want to find operators that can make this expression false for some truth values of \( p \) and \( q \).
### Step-by-Step Solution:
1. **Understanding the Expression**:
We have the expression \( p \, o+ \, (p \, ox \, q) \). We need to analyze under what conditions this expression can be false.
2. **Identifying Operators**:
We will consider the four options provided for \( o+ \) and \( ox \):
- Option 1: \( o+ \) is OR, \( ox \) is IMPLIES
- Option 2: \( o+ \) is IMPLIES, \( ox \) is AND
- Option 3: \( o+ \) is IMPLIES, \( ox \) is OR
- Option 4: None of these
3. **Testing Option 1**:
- Let \( o+ \) be OR and \( ox \) be IMPLIES.
- The expression becomes \( p \lor (p \implies q) \).
- This is true if \( p \) is true (regardless of \( q \)) or if \( p \) is false (because \( p \implies q \) is true when \( p \) is false).
- Thus, this expression is always true, and hence it is a tautology.
4. **Testing Option 2**:
- Let \( o+ \) be IMPLIES and \( ox \) be AND.
- The expression becomes \( p \implies (p \land q) \).
- This is false when \( p \) is true and \( q \) is false (since true implies false is false).
- Therefore, this expression can be false, making it not a tautology.
5. **Testing Option 3**:
- Let \( o+ \) be IMPLIES and \( ox \) be OR.
- The expression becomes \( p \implies (p \lor q) \).
- This is always true because if \( p \) is true, \( p \lor q \) is true, and if \( p \) is false, \( p \implies (p \lor q) \) is true.
- Thus, this expression is a tautology.
6. **Conclusion**:
From our analysis, only Option 2 satisfies the condition of not being a tautology. Therefore, \( o+ \) can be IMPLIES and \( ox \) can be AND.
### Final Answer:
The operators \( o+ \) and \( ox \) can be:
- \( o+ \): IMPLIES
- \( ox \): AND
