Chapter 4

Equivalence of Formulas and Normal Forms

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Problem 65

Prove that formula Ψ\Psi follows from formula Φ\Phi if and only if Φ∧Ψ≡Φ\Phi \wedge \Psi \equiv \Phi, if and only if Φ∨Ψ≡Ψ\Phi \vee \Psi \equiv \Psi.

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Problem 66

Verify all the equivalences 1)–9) given in Section 4.1 by directly constructing truth tables for the functions represented by their left- and right-hand sides.

The nine equivalences (with Φ\Phi, Ψ\Psi, Θ\Theta arbitrary formulas):

  1. Associativity: (Φ∧Ψ)∧Θ≡Φ∧(Ψ∧Θ)(\Phi \wedge \Psi ) \wedge \Theta \equiv \Phi \wedge (\Psi \wedge \Theta ); (Φ∨Ψ)∨Θ≡Φ∨(Ψ∨Θ)(\Phi \vee \Psi ) \vee \Theta \equiv \Phi \vee (\Psi \vee \Theta ); (Φ⊕Ψ)⊕Θ≡Φ⊕(Ψ⊕Θ)(\Phi \oplus \Psi ) \oplus \Theta \equiv \Phi \oplus (\Psi \oplus \Theta ).

  2. Commutativity: Φ∧Ψ≡Ψ∧Φ\Phi \wedge \Psi \equiv \Psi \wedge \Phi; Φ∨Ψ≡Ψ∨Φ\Phi \vee \Psi \equiv \Psi \vee \Phi; Φ⊕Ψ≡Ψ⊕Φ\Phi \oplus \Psi \equiv \Psi \oplus \Phi.

  3. Distributivity: (Φ∨Ψ)∧Θ≡(Φ∧Θ)∨(Ψ∧Θ)(\Phi \vee \Psi ) \wedge \Theta \equiv (\Phi \wedge \Theta ) \vee (\Psi \wedge \Theta ); (Φ∧Ψ)∨Θ≡(Φ∨Θ)∧(Ψ∨Θ)(\Phi \wedge \Psi ) \vee \Theta \equiv (\Phi \vee \Theta ) \wedge (\Psi \vee \Theta ); (Φ⊕Ψ)∧Θ≡(Φ∧Θ)⊕(Ψ∧Θ)(\Phi \oplus \Psi ) \wedge \Theta \equiv (\Phi \wedge \Theta ) \oplus (\Psi \wedge \Theta ).

  4. De Morgan’s laws: ¬(Φ∨Ψ)≡¬Φ∧¬Ψ\neg (\Phi \vee \Psi ) \equiv \neg \Phi \wedge \neg \Psi; ¬(Φ∧Ψ)≡¬Φ∨¬Ψ\neg (\Phi \wedge \Psi ) \equiv \neg \Phi \vee \neg \Psi.

  5. Double negation: ¬¬Φ≡Φ\neg \neg \Phi \equiv \Phi.

  6. Idempotence: Φ∧Φ≡Φ\Phi \wedge \Phi \equiv \Phi; Φ∨Φ≡Φ\Phi \vee \Phi \equiv \Phi.

  7. Laws for constants: Φ∧¬Φ≡0\Phi \wedge \neg \Phi \equiv 0; Φ∨¬Φ≡1\Phi \vee \neg \Phi \equiv 1; Φ∧0≡0\Phi \wedge 0 \equiv 0; Φ∨0≡Φ\Phi \vee 0 \equiv \Phi; Φ∧1≡Φ\Phi \wedge 1 \equiv \Phi; Φ∨1≡1\Phi \vee 1 \equiv 1.

  8. Φ→Ψ≡¬Φ∨Ψ\Phi \rightarrow \Psi \equiv \neg \Phi \vee \Psi.

  9. Φ⊕Ψ≡(Φ∧¬Ψ)∨(¬Φ∧Ψ)≡(Φ∨Ψ)∧(¬Φ∨¬Ψ)\Phi \oplus \Psi \equiv (\Phi \wedge \neg \Psi ) \vee (\neg \Phi \wedge \Psi ) \equiv (\Phi \vee \Psi ) \wedge (\neg \Phi \vee \neg \Psi ).

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Problem 67

Call a formula of the form Φ1∧Φ2∧…∧Φn\Phi_{1} \wedge \Phi_{2} \wedge \ldots \wedge \Phi_{n} a logical product. We call its subformulas Φi,1⩽i⩽n\Phi_{i}, 1 \leqslant i \leqslant n, the factors. Likewise, call a formula of the form Φ1∨Φ2∨…∨Φn\Phi_{1} \vee \Phi_{2} \vee \ldots \vee \Phi_{n} a logical sum. We call its subformulas Φi,1⩽i⩽n\Phi_{i}, 1 \leqslant i \leqslant n, the summands.

Show that the following rules for transforming logical products and sums can be derived from the basic identities:

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(a)

if at least one of the factors in a logical product equals 0, then the whole product equals 0;

(b)

if at least one of the summands in a logical sum equals 1, then the whole sum equals 1;

(c)

if in a logical product n⩾2n \geqslant 2 and there is a factor equal to 1, it can be crossed out;

(d)

if in a logical sum n⩾2n \geqslant 2 and there is a summand equal to 0, it can be crossed out.

Problem 68

Using the basic identities, prove the equivalence of the following pairs of formulas.

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(a)

¬(x∨¬y)∧(x→¬y)\neg (x \vee \neg y) \wedge (x \rightarrow \neg y) and ¬x∧y\neg x \wedge y;

(b)

¬((x∧¬y)→(¬x∨z))\neg ((x \wedge \neg y) \rightarrow (\neg x \vee z)) and x∧¬y∧¬zx \wedge \neg y \wedge \neg z;

(c)

(x⊕y)→(x∧¬y)(x \oplus y) \rightarrow (x \wedge \neg y) and (¬x∧¬y)∨x(\neg x \wedge \neg y) \vee x.

Problem 69

Let the interpretation JJ differ from the interpretation II only in that J(x)=I(Θ)J(x)=I(\Theta ). By induction on the construction of the formula Φ\Phi, prove that J(Φ)=I((Φ)Θx)J(\Phi )=I\left((\Phi )_{\Theta }^{x}\right).

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Problem 70

Prove Proposition 19 on p. 81.

Proposition 19: If in two equivalent formulas Φ\Phi and Ψ\Psi the propositional variable xx is replaced by a formula Θ\Theta, then the resulting formulas are again equivalent: if Φ≡Ψ\Phi \equiv \Psi, then (Φ)Θx≡(Ψ)Θx(\Phi )_{\Theta }^{x} \equiv (\Psi )_{\Theta }^{x}.

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Problem 71

A Boolean function f∗(x1,…,xn)f^{*}\left(x_{1}, \ldots , x_{n}\right) is called dual to the function f(x1,…,xn)f\left(x_{1}, \ldots , x_{n}\right) if f∗(x1,…,xn)=¬f(¬x1,…,¬xn)f^{*}\left(x_{1}, \ldots , x_{n}\right)=\neg f\left(\neg x_{1}, \ldots , \neg x_{n}\right) for every assignment of values to the variables. For example, conjunction is dual to disjunction and vice versa.

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(a)

Prove that the duality relation is symmetric.

(b)

Let f(y1,…,ym),g1(x1,…,xn),…,gm(x1,…,xn)f\left(y_{1}, \ldots , y_{m}\right), g_{1}\left(x_{1}, \ldots , x_{n}\right), \ldots , g_{m}\left(x_{1}, \ldots , x_{n}\right) be Boolean functions and

h(x1,…,xn)=f(g1(x1,…,xn),…,gm(x1,…,xn)). h\left(x_{1}, \ldots , x_{n}\right)=f\left(g_{1}\left(x_{1}, \ldots , x_{n}\right), \ldots , g_{m}\left(x_{1}, \ldots , x_{n}\right)\right).

Establish the following duality principle: the dual function of a superposition of functions equals the superposition of the dual functions:

h∗(x1,…,xn)=f∗(g1∗(x1,…,xn),…,gm∗(x1,…,xn)). h^{*}\left(x_{1}, \ldots , x_{n}\right)=f^{*}\left(g_{1}^{*}\left(x_{1}, \ldots , x_{n}\right), \ldots , g_{m}^{*}\left(x_{1}, \ldots , x_{n}\right)\right).
Problem 72

By induction on the construction of the formula Φ\Phi, prove the inequality

depth⁡(Φ)Ψx⩽depth⁡Φ+depth⁡Ψ \operatorname {depth}(\Phi )_{\Psi }^{x} \leqslant \operatorname {depth} \Phi +\operatorname {depth} \Psi

Give an example where the inequality is strict.

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Problem 73

Recall the notation: for a formula Φ\Phi and σ∈{0,1}\sigma \in \left\{ 0,1\right\}, write Φσ=Φ\Phi^{\sigma } = \Phi if σ=1\sigma = 1 and Φσ=¬Φ\Phi^{\sigma } = \neg \Phi if σ=0\sigma = 0; consequently 1σ≡σ1^{\sigma } \equiv \sigma, 0σ≡¬σ0^{\sigma } \equiv \neg \sigma, σσ≡1\sigma^{\sigma } \equiv 1, and στ≡0\sigma^{\tau } \equiv 0 for σ≠τ\sigma \neq \tau (formula (7)). For a Boolean function f(x1,…,xn)f\left(x_1, \ldots , x_n\right) define

Df=⋁f(σ1,…,σn)=1⋀i=1nxiσi,Cf=⋀f(σ1,…,σn)=0⋁i=1nxi¬σi. \mathcal{D}_{f} = \bigvee _{f\left(\sigma _{1}, \ldots , \sigma _{n}\right)=1} \bigwedge _{i=1}^{n} x_{i}^{\sigma _{i}}, \qquad \mathcal{C}_{f} = \bigwedge _{f\left(\sigma _{1}, \ldots , \sigma _{n}\right)=0} \bigvee _{i=1}^{n} x_{i}^{\neg \sigma _{i}}.

Theorem 21: Let ff be a Boolean function. 1) If ff is not identically zero, then Df\mathcal{D}_{f} is a perfect DNF representing ff. 2) If ff is not identically one, then Cf\mathcal{C}_{f} is a perfect CNF representing ff. (Part 1 is proved in the text; part 2 is this problem.)

Prove the second part of Theorem 21 on p. 84: for every assignment of argument values σ1,…,σn\sigma_{1}, \ldots , \sigma_{n} we have f(σ1,…,σn)=Cf(σ1,…,σn)f\left(\sigma_{1}, \ldots , \sigma_{n}\right)=\mathcal{C}_{f}\left(\sigma_{1}, \ldots , \sigma_{n}\right).

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Problem 74

Propose procedures for solving the following problems:

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(a)

Given an arbitrary elementary conjunction, construct an equivalent perfect DNF over a given set of variables.

(b)

Given an arbitrary elementary disjunction, construct an equivalent perfect CNF over a given set of variables.

Problem 75

Prove that for all k⩽nk \leqslant n every Boolean function f∈Pnf \in \mathcal{P}_{n} can be represented as

f(x1,…,xk,xk+1,…,xn)==⋁(σ1,…,σk)∈Bkx1σ1∧…∧xkσk∧f(σ1,…,σk,xk+1,…,xn). \begin{aligned} f(x_{1}, \ldots , x_{k}, x_{k+1}, \ldots , x_{n})= \\ & =\bigvee _{\left(\sigma _{1}, \ldots , \sigma _{k}\right) \in \mathbb {B}^{k}} x_{1}^{\sigma _{1}} \wedge \ldots \wedge x_{k}^{\sigma _{k}} \wedge f(\sigma _{1}, \ldots , \sigma _{k}, x_{k+1}, \ldots , x_{n}). \end{aligned}

Such a representation is called the expansion of ff in x1,…,xkx_{1}, \ldots , x_{k}. For k=nk=n it yields the perfect DNF from Theorem 21 on p. 84.

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Problem 76

How should the procedure for reducing to a perfect DNF be changed so as to obtain a procedure for reducing to a perfect CNF?

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Problem 77

Propose a method for simultaneously constructing equivalent DNF and CNF for an arbitrary formula Φ\Phi of propositional logic, using induction on the construction of Φ\Phi.

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Problem 78

Find equivalent reduced DNFs and prove the equivalence of the following pairs of formulas:

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(a)

Φ=(((¬x∧¬y)→¬z)∧(x→y)),Ψ=((1⊕y)→(¬x∧(1⊕z)))\Phi =(((\neg x \wedge \neg y) \rightarrow \neg z) \wedge (x \rightarrow y)), \Psi =((1 \oplus y) \rightarrow (\neg x \wedge (1 \oplus z)));

(b)

Φ=(¬((x1→x2)∨¬(x2→x1))∧x3),Ψ=¬((x1∧x3)→x2)\Phi =\left(\neg \left(\left(x_{1} \rightarrow x_{2}\right) \vee \neg \left(x_{2} \rightarrow x_{1}\right)\right) \wedge x_{3}\right), \Psi =\neg \left(\left(x_{1} \wedge x_{3}\right) \rightarrow x_{2}\right);

(c)

Φ=¬(¬x∧y∧¬z)→((y⊕1)∧((x⊕1)→¬(¬u∨z)))\Phi =\neg (\neg x \wedge y \wedge \neg z) \rightarrow ((y \oplus 1) \wedge ((x \oplus 1) \rightarrow \neg (\neg u \vee z))), Ψ=(¬x∨y)→((¬u∨y∨z)→(¬(x∨¬y)∧¬z));\Psi =(\neg x \vee y) \rightarrow ((\neg u \vee y \vee z) \rightarrow (\neg (x \vee \neg y) \wedge \neg z)) ;

(d)

Φ=(¬(x→(¬y→(x∧¬z)))∧(z∨¬(x∧y)))\Phi =(\neg (x \rightarrow (\neg y \rightarrow (x \wedge \neg z))) \wedge (z \vee \neg (x \wedge y))), Ψ=((x∧z)⊕(x∧y∧z));\Psi =((x \wedge z) \oplus (x \wedge y \wedge z)) ;

(e)

Φ=(((x∧y)→¬z)∧(¬x→¬y)),Ψ=(y→(x∧(z⊕1)))\Phi =(((x \wedge y) \rightarrow \neg z) \wedge (\neg x \rightarrow \neg y)), \Psi =(y \rightarrow (x \wedge (z \oplus 1)));

(f)

Φ=(((x∨y)→¬z)∧((x∧z)→y)),Ψ=(z→((x⊕1)∧¬y))\Phi =(((x \vee y) \rightarrow \neg z) \wedge ((x \wedge z) \rightarrow y)), \Psi =(z \rightarrow ((x \oplus 1) \wedge \neg y)).

Problem 79

Suppose a DNF D\mathcal{D} contains no negations and the absorption laws cannot be applied to it. Prove that D\mathcal{D} is a reduced DNF.

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Problem 80

Prove the equivalences (8)–(11).

  1. ¬x≡(x⊕1)\neg x \equiv (x \oplus 1).

  2. (x1∧x2)≡(x1x2)\left(x_{1} \wedge x_{2}\right) \equiv \left(x_{1} x_{2}\right).

  3. (x1∨x2)≡(x1x2⊕x1⊕x2)\left(x_{1} \vee x_{2}\right) \equiv \left(x_{1} x_{2} \oplus x_{1} \oplus x_{2}\right).

  4. (x1⊕x2)(x3⊕x4)≡(x1x3⊕x1x4⊕x2x3⊕x2x4)\left(x_{1} \oplus x_{2}\right)\left(x_{3} \oplus x_{4}\right) \equiv \left(x_{1} x_{3} \oplus x_{1} x_{4} \oplus x_{2} x_{3} \oplus x_{2} x_{4}\right).

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Problem 81

Prove equalities (12) and (13).

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Problem 82

Using the basic equivalences and identities (8)–(11), find the equivalent reduced Zhegalkin polynomials and prove the equivalence of the following pairs of formulas:

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(a)

Φ=((z∧(x→y))∨¬(¬x→z)),Ψ=(x→(y∧z))\Phi =((z \wedge (x \rightarrow y)) \vee \neg (\neg x \rightarrow z)), \Psi =(x \rightarrow (y \wedge z));

(b)

Φ=¬(x→(y∧z))∧(¬y∨¬x)\Phi =\neg (x \rightarrow (y \wedge z)) \wedge (\neg y \vee \neg x), Ψ=(¬(x→y)∨z)∧¬((¬x∧z)∨(y∧z));\Psi =(\neg (x \rightarrow y) \vee z) \wedge \neg ((\neg x \wedge z) \vee (y \wedge z)) ;

(c)

Φ=(((y∧z)→¬(x∨z))∧¬(¬y∧z∧x)),Ψ=(z→(¬y∧¬x))\Phi =(((y \wedge z) \rightarrow \neg (x \vee z)) \wedge \neg (\neg y \wedge z \wedge x)), \Psi =(z \rightarrow (\neg y \wedge \neg x));

(d)

Φ=¬(((x→y)∨¬z)∧x),Ψ=(¬(z→y)∨¬x)\Phi =\neg (((x \rightarrow y) \vee \neg z) \wedge x), \Psi =(\neg (z \rightarrow y) \vee \neg x);

(e)

Φ=(¬((x→y)∨¬(y→x))∧z),Ψ=¬((x∧z)→y)\Phi =(\neg ((x \rightarrow y) \vee \neg (y \rightarrow x)) \wedge z), \Psi =\neg ((x \wedge z) \rightarrow y).

Problem 83

Find the Zhegalkin polynomials (by the method of undetermined coefficients and using the matrix JJ) for the following functions f(x,y,z)f(x, y, z). Assume that the argument tuples of the functions are ordered lexicographically and their values are given by a sequence of 8 zeros and ones:

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(a)

f=(00101100)f=(00101100);

(b)

f=(11101100)f=(11101100);

(c)

f=(11000011)f=(11000011);

(d)

f=(01101011)f=(01101011).

Problem 84

Find the Boolean function of nn variables whose reduced Zhegalkin polynomial has the greatest length.

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