Key Concepts and Formulas

• Implication: $A \to B \equiv \sim A \vee B$
• Tautology: A statement that is always true, regardless of the truth values of its components.
• De Morgan's Laws: $\sim (A \wedge B) \equiv \sim A \vee \sim B$ and $\sim (A \vee B) \equiv \sim A \wedge \sim B$
• Commutative Laws: $A \wedge B \equiv B \wedge A$ and $A \vee B \equiv B \vee A$
• Associative Laws: $(A \wedge B) \wedge C \equiv A \wedge (B \wedge C)$ and $(A \vee B) \vee C \equiv A \vee (B \vee C)$
• Distributive Laws: $A \wedge (B \vee C) \equiv (A \wedge B) \vee (A \wedge C)$ and $A \vee (B \wedge C) \equiv (A \vee B) \wedge (A \vee C)$
• Identity Laws: $A \wedge T \equiv A$ and $A \vee F \equiv A$ where T is always True and F is always False.
• Complement Laws: $A \wedge \sim A \equiv F$ and $A \vee \sim A \equiv T$
• Absorption Laws: $A \wedge (A \vee B) \equiv A$ and $A \vee (A \wedge B) \equiv A$

Step-by-Step Solution

Step 1: Analyze option (A): $B \to \left[ {A \wedge \left( {A \to B} \right)} \right]$

We need to check if $B \to \left[ {A \wedge \left( {A \to B} \right)} \right]$ is a tautology. Let's rewrite the implication:

$B \to \left[ {A \wedge \left( {\sim A \vee B} \right)} \right] \equiv \sim B \vee \left[ {A \wedge \left( {\sim A \vee B} \right)} \right]$

Applying the distributive law:

$\sim B \vee \left[ {\left( {A \wedge \sim A} \right) \vee \left( {A \wedge B} \right)} \right] \equiv \sim B \vee \left[ {F \vee \left( {A \wedge B} \right)} \right] \equiv \sim B \vee \left( {A \wedge B} \right)$

$\equiv \left( {\sim B \vee A} \right) \wedge \left( {\sim B \vee B} \right) \equiv \left( {\sim B \vee A} \right) \wedge T \equiv \sim B \vee A$

This is not a tautology. The truth value depends on A and B.

Step 2: Analyze option (B): $\left[ {A \wedge \left( {A \to B} \right)} \right] \to B$

We need to check if $\left[ {A \wedge \left( {A \to B} \right)} \right] \to B$ is a tautology. Let's rewrite the implication:

$\left[ {A \wedge \left( {A \to B} \right)} \right] \to B \equiv \left[ {A \wedge \left( {\sim A \vee B} \right)} \right] \to B$

Using the implication rule again:

$\sim \left[ {A \wedge \left( {\sim A \vee B} \right)} \right] \vee B \equiv \sim A \vee \sim \left( {\sim A \vee B} \right) \vee B$

Applying De Morgan's Law:

$\sim A \vee \left( {A \wedge \sim B} \right) \vee B \equiv \left( {\sim A \vee A} \right) \wedge \left( {\sim A \vee \sim B} \right) \vee B \equiv T \wedge \left( {\sim A \vee \sim B} \right) \vee B$

$\equiv \left( {\sim A \vee \sim B} \right) \vee B \equiv \sim A \vee \left( {\sim B \vee B} \right) \equiv \sim A \vee T \equiv T$

Therefore, $\left[ {A \wedge \left( {A \to B} \right)} \right] \to B$ is a tautology.

Step 3: Analyze option (C): $\left[ {A \wedge \left( {A \vee B} \right)} \right]$

Using the absorption law: $A \wedge (A \vee B) \equiv A$. This is just $A$, which is not a tautology.

Step 4: Analyze option (D): $\left[ {A \vee \left( {A \wedge B} \right)} \right]$

Using the absorption law: $A \vee (A \wedge B) \equiv A$. This is just $A$, which is not a tautology.

Common Mistakes & Tips

• Be careful with the order of operations when applying logical equivalences.
• Remember the truth tables for basic logical connectives like AND, OR, NOT, and implication.
• Use the absorption laws to simplify expressions quickly.

Summary

By analyzing each option and simplifying the logical expressions using equivalences, we found that the statement $\left[ {A \wedge \left( {A \to B} \right)} \right] \to B$ is a tautology, as it simplifies to $T$.

Final Answer

The final answer is \boxed{B}, which corresponds to option (B).