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Write pseudocode for a program that, given a formula \(\phi,\) finds (i) a logically equivalent formula \(\phi^{\prime}\) in CNF and (ii) a logically equivalent formula \(\phi^{\prime \prime}\) in DNF. The algorithm should be recursive (similar to an induction on formulas) and should not involve the construction of truth tables. Prove the algorithm works. This gives an alternate proof of the theorem that every formula is equivalent to a formula in CNF.

Short Answer

Expert verified
Use recursive conversion for CNF and DNF, leveraging laws like De Morgan's and distributive properties.

Step by step solution

01

Base Case for Atoms

For the simplest formulas such as propositional variables (atoms), they are already in both CNF and DNF. The CNF and DNF of an atom \(p\) are both just \(p\) itself. In pseudocode,```function CNF(p): if p is an atomic formula: return p```Similarly, for DNF,```function DNF(p): if p is an atomic formula: return p```
02

Recursive Case for Negation

For a negation \(eg \phi\), we can use De Morgan's laws recursively. Convert \(\phi\) to CNF (or DNF), then apply the laws:```function CNF(eg \phi): return DNF( convert \phi to DNF apply De Morgan's laws )``````function DNF(eg \phi): return CNF( convert \phi to CNF apply De Morgan's laws )```
03

Recursive Case for Disjunction

For a disjunction \(\phi_1 \lor \phi_2\), the CNF can be found by applying the distributive laws. Convert \(\phi_1\) and \(\phi_2\) recursively:```function CNF(\phi_1 \lor \phi_2): convert \phi_1 and \phi_2 to CNF apply distributive laws if needed return result```In a DNF form, a disjunction is just a union of the converted DNFs:```function DNF(\phi_1 \lor \phi_2): return (convert \phi_1 to DNF) \lor (convert \phi_2 to DNF)```
04

Recursive Case for Conjunction

For a conjunction \(\phi_1 \land \phi_2\), a CNF is a simple conjunction of the converted CNFs:```function CNF(\phi_1 \land \phi_2): return (convert \phi_1 to CNF) \land (convert \phi_2 to CNF)```For DNF, apply distributive laws if necessary:```function DNF(\phi_1 \land \phi_2): convert \phi_1 and \phi_2 to DNF apply distributive laws if needed return result```
05

Algorithm Proof

This recursive algorithm works as it respects the structural induction on the formula. By correctly converting atomic formulas, handling negations using De Morgan’s laws, and using distributive laws correctly on conjunctions and disjunctions, we transform any formula into CNF or DNF. This systematically ensures the formula respects logical equivalency transformations as recognized by logical equivalences.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logical Equivalence
Logical equivalence is a fundamental concept in logic and mathematics. It refers to two logical statements or formulas that always have the same truth value in every possible situation.
Understanding logical equivalence is crucial when transforming formulas into CNF (Conjunctive Normal Form) or DNF (Disjunctive Normal Form), as we aim to create an equivalent formula that is easier to manipulate or analyze.
To identify logical equivalence, you do not need to construct truth tables. Instead, understanding and applying transformation rules like De Morgan’s laws or the distributive laws can achieve this equivalence transformation.
For instance, if you have two statements, say \(A o B\) and \(eg A \lor B\), they are logically equivalent and prove it without a truth table. Logical equivalence allows these transformations to be seamlessly integrated into algorithms for CNF and DNF conversion.
De Morgan's Laws
De Morgan's laws are essential rules in logic. They describe how negation interacts with conjunctions and disjunctions, allowing you to rewrite logical statements in a different yet equivalent form.
  • \(eg(A \land B)\) is equivalent to \(eg A \lor eg B\)
  • \(eg(A \lor B)\) is equivalent to \(eg A \land eg B\)
Using De Morgan's laws is crucial when transforming formulas involving negations into CNF or DNF.
By applying these laws, you can ensure that negations are pushed down to the atomic level, which simplifies further transformation processes.
Within the pseudocode for converting formulas to CNF or DNF, whenever a negation operator is encountered, De Morgan's laws guide the manner in which the formula will be broken down and restructured, maintaining logical equivalence.
Distributive Laws
The distributive laws in logic refer to the ability to distribute logical connectives over each other, similar to how multiplication distributes over addition in algebra.
  • Conjunction distributes over disjunction: \((A \land (B \lor C))\) is equivalent to \((A \land B) \lor (A \land C)\)
  • Disjunction distributes over conjunction: \((A \lor (B \land C))\) is equivalent to \((A \lor B) \land (A \lor C)\)
These laws are pivotal when converting complex formulas to CNF or DNF. For CNF, conjunctions need to distribute over disjunctions, ensuring the formula is a conjunction of disjunctions.
Conversely, for DNF, disjunctions distribute over conjunctions, presenting the formula as a disjunction of conjunctions.
Applying these laws within the pseudocode transforms the logical structure while retaining its truthfulness and sets the stage for further logical manipulation.
Structural Induction
Structural induction is a form of mathematical induction used to prove the properties of recursively defined structures, like logical formulas.
In the context of CNF and DNF conversion, structural induction validates that every step of the transformation maintains logical equivalence.
The process begins with the simplest base case—atomic formulas—which are inherently in CNF and DNF.
  • Base Case: Atomic formulas like \(p\) are in both CNF and DNF.
  • Induction Step: Assuming formulas \(\phi_1\) and \(\phi_2\) can be converted to CNF and DNF, show that combined formulas \(eg\phi_1\), \(\phi_1 \land \phi_2\), and \(\phi_1 \lor \phi_2\) can also be converted.
By using structural induction, the recursive algorithm for CNF and DNF conversion is proven reliable. It confirms each logical operator's transformation aligns with known logical laws, ensuring the end result is equivalent to the original formula.

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Most popular questions from this chapter

Let the universe \(U\) be the set of all human beings living in the year \(2001,\) and translate the following English sentences into quantified formulas. Let \(P(x)\) stand for \(" x\) is young" \(Q(x)\) for \(^{4} x\) is female," and \(R(x)\) for \(^{*} x\) is an athlete." (a) "All athletes are young." (b) "Not all young people are athletes." (c) "All young people are not athletes." (Warning: In informal English, this sentence has two quite different meanings. One is "more grammatically correct" than the other, however, and that is the one we're asking for.) (d) "Some young people are not athletes." (e) "Some athletes are young females." (f) "All athletes are young males." (g) "Some athletes are female and are not young." (h) "Some young females are not athletes." (i) "All young females are athletes." (j) "Some athletes are not young." (k) "No young people are athletes." (1) "All athletes are either female or are young." (m) "If all athletes are female, then all athletes are young; otherwise, no athletes are young."

Find the expression tree for the formula $$ ((p \rightarrow \neg q) \vee q) \rightarrow q $$ Evaluate the expression tree if proposition \(p\) is \(F\) and proposition \(q\) is \(T\).

(a) Show that \((p \vee q)\) is an alphabetic variant of \((q \vee p)\). (b) Show that the relation of being an alphabetic variant is an equivalence relation. (c) Show that if \(\psi\) is an alphabetic variant of \(\phi .\) then \(\phi\) is a tautology (respectively, is satisfiable, is unsatisfiable) if and only if \(\psi\) is a tautology (respectively, is satisfiable, is unsatisfiable). (d) Show that \(\phi\) being an alphabetic variant of \(\psi\) does not imply that \(\phi\) and \(\psi\) are tautologically equivalent.

(a) Show that every formula containing only \(k\) (different) proposition letters is equivalent to a \(k-\mathrm{DNF}\) formula. (b) Show that \(p \leftrightarrow q\) is not equivalent to any 1 -DNF formula. (c) Show that for every natural number \(k\) (including 0 ), there is a formula containing only \(k+1\) (different) proposition letters that is not equivalent to any \(k-D N F\) formula.

(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 \(n\).)

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