Question

Let \(\Delta, \nabla \in\{\Lambda, V\}\) be such that \(( p \rightarrow q ) \Delta( p \nabla q )\) is a tautology. Then

• A. $\Delta=\Lambda, \nabla=\Lambda$
• B. $\Delta=V, \nabla=V$
• C. $\Delta=V, \nabla=\wedge$
• D. $\Delta=\Lambda, \nabla=V$

Answer 1

The Correct Option is B

To solve the given problem, we must determine the correct logical operations for \( \Delta \) and \( \nabla \) such that the expression \( (p \rightarrow q) \Delta (p \nabla q) \) is a tautology. A tautology is a statement that is always true, regardless of the truth values of its components.

Let's analyze the expression \( (p \rightarrow q) \Delta (p \nabla q) \):

1. \(p \rightarrow q\) is a logical implication. It is false only when \( p \) is true and \( q \) is false; otherwise, it is true.
2. We need to check which of the given combinations of \( \Delta \) and \( \nabla \) result in the entire expression being a tautology.

Now, evaluate the possibilities:

1. If \(\nabla = \wedge\), then \( p \nabla q = p \wedge q \). This is true only if both \( p \) and \( q \) are true.
2. If \(\nabla = V\) (logical OR), then \( p \nabla q = p \vee q \). This is true if either \( p \) or \( q \) is true.

We need the expression \( (p \rightarrow q) \Delta (p \nabla q) \) to be always true:

1. Consider \( \Delta = V\). Then \( (p \rightarrow q) V (p \nabla q) \) translates to \((p \rightarrow q) \vee (p \nabla q)\).
   • If \( \nabla = \wedge\), then it must be checked for various truth values:
     • For \( p = \text{true}, q = \text{true} \): \((\text{true} \rightarrow \text{true} ) \vee (\text{true} \wedge \text{true}) = \text{true} \vee \text{true} = \text{true}\).
     • For \( p = \text{true}, q = \text{false} \): \((\text{true} \rightarrow \text{false}) \vee (\text{true} \wedge \text{false}) = \text{false} \vee \text{false} = \text{false}\) (so not always true).
   • If \( \nabla = V\), this makes \( p \nabla q = p \vee q \), which results in:
     • For all truth values of \( p \) and \( q \), the expression is evaluated to true as \((p \rightarrow q) \vee (p \vee q)\) is equivalent to the tautology.
2. Now with \(\Delta = \vee\) and \(\nabla = \vee\), the statement is always true and is thus a tautology.

Therefore, the correct logical connectives are \(\Delta = V\) and \(\nabla = V\). The statement \( (p \rightarrow q) \vee (p \vee q) \) is a tautology as it is true for all possible truth values.