Question

(a) The conjunction of \(n\) formulas \(p_{1}, p_{2}, \ldots, p_{n}\) is defined to be the formula \(\left(\ldots\left(\left(p_{1} \wedge p_{2}\right) \wedge p_{3}\right) \wedge \ldots\right) \wedge p_{n} .\) For \(n=0,\) there is a special case: The conjunction of zero formulas is defined to be \(T\). For \(n=1\), that conjunction simplifies to \(p_{1}\). Let \(\phi\) be the conjunction of \(p_{1}, p_{2}, \ldots, p_{n} .\) Prove that for any interpretation \(I, I(\phi)=T\) if and only if \(I\left(p_{i}\right)=T\) for each \(i\) such that \(1 \leq i \leq n .\) (Hint: Use induction.) (b) Let \(\phi\) be the formula $$ \left(\ldots\left(\left(p_{1} \leftrightarrow p_{2}\right) \leftrightarrow p_{3}\right) \leftrightarrow \ldots\right) \leftrightarrow p_{n} $$ for \(n \geq 1 .\) For what interpretations \(I\) is \(I(\phi)=T ?\) (Hint: The answer involves counting how many of the \(p_{i}\) 's are true in \(I\). Prove the result by induction on \(n\).)

Short answer

(a) The conjunction is true when all \( p_i \) are true. (b) The biconditional formula is true for even true literals.

Step-by-step solution

Understanding the base case for conjunction

According to the problem, when \( n = 0 \), the conjunction of zero formulas is defined to be \( T \). For \( n = 1 \), the conjunction simplifies to \( p_1 \). For the base case, we need to check if the conjunction outcomes match the given initial condition: if \( n = 1 \), then \( I(\phi) = T \) if and only if \( I(p_1) = T \), which holds true by definition.

Inductive Hypothesis for conjunction

Assume for some \( k \geq 1 \), if \( I(\phi_k) = T \), then \( I(p_i) = T \) for every \( 1 \leq i \leq k \). \( \phi_k \) represents the conjunction of \( p_1, p_2, \dots, p_k \). This will be our inductive hypothesis.

Inductive Step for conjunction

We now show that if \( I(\phi_k) = T \), then it holds for \( k+1 \). The formula \( \phi_{k+1} \) is given by \( (\phi_k \wedge p_{k+1}) \). For \( I(\phi_{k+1}) = T \), it requires \( I(\phi_k) = T \) and \( I(p_{k+1}) = T \). By the inductive hypothesis, \( I(p_i) = T \) for all \( 1 \leq i \leq k \). Hence, \( I(p_i) = T \) for all \( 1 \leq i \leq k+1 \). Thus, the statement holds for \( k+1 \), completing the inductive step.

Base Case for biconditional formula

Consider the base case when \( n = 1 \). Here, \( \phi = p_1 \), and obviously, \( I(\phi) = T \) if and only if \( I(p_1) = T \).

Inductive Hypothesis for biconditional formula

Assume that for \( n = m \), the formula \( \phi_m = (\dots((p_1 \leftrightarrow p_2) \leftrightarrow p_3) \dots \leftrightarrow p_m) \) is true if and only if an even number of \( p_i \)'s (where \( 1 \leq i \leq m \)) are true.

Inductive Step for biconditional formula

To show the hypothesis holds for \( n = m+1 \), consider \( \phi_{m+1} = (\phi_m \leftrightarrow p_{m+1}) \). The expression \( (A \leftrightarrow B) \) is true if both are equal. For even number true at \( \phi_m \): - If \( I(p_{m+1}) = T \), then the count becomes odd, hence \( \phi_{m+1} = F \).- If \( I(p_{m+1}) = F \), then the count remains even, hence \( \phi_{m+1} = T \).Thus, the evenness or oddness switches with each successive \( I(p_i) \). This shows the parity determines \( I(\phi_{m+1}) = T \).

Conclusion on biconditional formula

From the inductive step, \( I(\phi) = T \) if and only if there is an even number of true \( p_i \)'s for \( n \geq 1 \). Each additional biconditional operation respects the parity of true values, reaffirming the hypothesis.

Key concepts

Conjunction

In logic, conjunction is a fundamental operation represented by the symbol \( \wedge \). It combines multiple statements into a single, composite statement.
For instance, the conjunction of statements \( p_1, p_2, \ldots, p_n \) can be visualized as cascading logical AND operations: \( ((p_1 \wedge p_2) \wedge p_3) \wedge \ldots \wedge p_n \).
This operation results in a true statement if and only if every individual component statement \( p_1, p_2, \ldots, p_n \) is true.
### Key Points on Conjunction:

• If any component of the conjunction is false, the entire conjunction becomes false.
• For \( n=0 \), the conjunction of zero statements is defined as true (\( T \)), which may seem counterintuitive, but serves to preserve logical consistency. This is akin to the concept of the empty product in arithmetic.
• The conjunction operation highlights how interdependent logical statements can create comprehensive logical expressions.

Using mathematical induction provides a powerful way to prove the properties of conjunction for any number \( n \) of statements.

Biconditional

The biconditional logical operator is represented by \( \leftrightarrow \) and is a way to express equivalence between two statements.
It's true when both statements have the same truth value, i.e., when both are true or both are false. Specifically, a formula composed as \( ((p_1 \leftrightarrow p_2) \leftrightarrow p_3) \leftrightarrow \ldots \leftrightarrow p_n \) checks for a consistent pattern of truth or falsity.
### Key Properties of Biconditional:

• When all involved statements are true or all are false, the resultant biconditional expression is true.
• The biconditional checks parity in a sequence of truths and observantly tracks parity flips as it processes each consequent operator.
• This operation can be used in proofs requiring precise logical equivalences.

An inductive approach reveals the condition under which such complex biconditional structures yield a true value—specifically, that an even number of statements are true among any finite collection \( p_i \). This intrinsic sensitivity to evenness or oddness plays a crucial role in constructing logical proofs involving such operators.

Truth Assignment

In propositional logic, a truth assignment is a method of determining the truth or falsity of each variable within a logical expression.
It's a vital concept used to assess the outcome of logical expressions under various conditions. Each variable within a formula can either be assigned true (\( T \)) or false (\( F \)), and these assignments can significantly change the result of complex logical constructs.
### How Truth Assignment Works:

• Truth assignments are foundational, as they provide the basis for evaluating logical statements.
• Each formula under consideration is interpreted according to the given truth assignment, which determines the overall truth value of the formula.
• For any induction on logical expressions, such as conjunctions or biconditionals, truth assignments are key to comparing initial assumptions with new outcomes.

A deep understanding of truth assignments is crucial for working effectively with logical expressions, enabling a deeper comprehension of deductive reasoning and the mechanics underlying logical proofs.