Let `o+and ox` are two mathematical operators . If `po+(q oxr)` is equivalent to `((p^^q)rArrr)` , then `o+ and ox`

Options

A. can be `vv and ^^` respectively
B. can be `^^ and vv` respectively
C. can both be `rArr`
D. can be `rArr and hArr` respectively

Answer

To solve the problem, we need to find the mathematical operators \( o+ \) and \( o_x \) such that the expression \( p o+ (q o_x r) \) is equivalent to \( (p \land q) \Rightarrow r \).

### Step 1: Understand the Logical Operators
We need to identify the logical operators that can replace \( o+ \) and \( o_x \). The logical operators we commonly use are:
- Conjunction (\( \land \)): AND
- Disjunction (\( \lor \)): OR
- Implication (\( \Rightarrow \)): If...then
- Biconditional (\( \Leftrightarrow \)): If and only if

### Step 2: Create the Truth Table for \( (p \land q) \Rightarrow r \)
The expression \( (p \land q) \Rightarrow r \) can be analyzed using a truth table:

| p | q | r | p ∧ q | (p ∧ q) ⇒ r |
|-------|-------|-------|-------|-------------|
| T | T | T | T | T |
| T | T | F | T | F |
| T | F | T | F | T |
| T | F | F | F | T |
| F | T | T | F | T |
| F | T | F | F | T |
| F | F | T | F | T |
| F | F | F | F | T |

From the truth table, we can see that \( (p \land q) \Rightarrow r \) is false only when \( p \) and \( q \) are true and \( r \) is false. In all other cases, it is true.

### Step 3: Analyze the Expression \( p o+ (q o_x r) \)
We need to determine the operators \( o+ \) and \( o_x \) such that the expression \( p o+ (q o_x r) \) yields the same truth values as \( (p \land q) \Rightarrow r \).

### Step 4: Testing Possible Operators
We will test various combinations of logical operators for \( o+ \) and \( o_x \).

1. **Option 1**: Let \( o+ \) be Disjunction (\( \lor \)) and \( o_x \) be Conjunction (\( \land \)).
- Expression: \( p \lor (q \land r) \)
- Truth Table:

| p | q | r | q ∧ r | p ∨ (q ∧ r) |
|-------|-------|-------|-------|-------------|
| T | T | T | T | T |
| T | T | F | F | T |
| T | F | T | F | T |
| T | F | F | F | T |
| F | T | T | T | T |
| F | T | F | F | F |
| F | F | T | F | T |
| F | F | F | F | F |

This does not match the original expression.

2. **Option 2**: Let \( o+ \) be Conjunction (\( \land \)) and \( o_x \) be Disjunction (\( \lor \)).
- Expression: \( p \land (q \lor r) \)
- Truth Table:

| p | q | r | q ∨ r | p ∧ (q ∨ r) |
|-------|-------|-------|-------|-------------|
| T | T | T | T | T |
| T | T | F | T | T |
| T | F | T | T | T |
| T | F | F | F | F |
| F | T | T | T | F |
| F | T | F | T | F |
| F | F | T | T | F |
| F | F | F | F | F |

This does not match the original expression.

3. **Option 3**: Let \( o+ \) and \( o_x \) both be Implication (\( \Rightarrow \)).
- Expression: \( p \Rightarrow (q \Rightarrow r) \)
- Truth Table:

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

This matches the original expression.

### Conclusion
The operators \( o+ \) and \( o_x \) are both Implication (\( \Rightarrow \)).