Using the Equivalence Laws

We can be establish an equivalence in a sequence of steps that create a chain:

A1 ≡ A² ≡ . . . ≡ Aⁿ

This, via transitivity, proves: A1 ≡ Aⁿ. The idea is that each of the equivalences Ai ≡ Aⁱ⁺¹ should be easily deducible, or previously established. Here is an example:

¬[(A∧ B) ∨¬C] ≡ ¬(A ∧B) ∧¬¬C ≡ ¬(A ∧ B) ∧C ≡ (¬A ∨ ¬B) ∧C ≡ (¬A ∧ C) ∨ (¬B ∧ C) Find for yourself what law (or laws) is used in each step. The chain proves:

¬[(A ∧ B) ∨ ¬C] ≡ (¬A ∧ C) ∨ (¬B ∧ C) . In this way we have simplified

¬[(A ∧ B) ∨ ¬C] to (¬A ∧ C) ∨ (¬B ∧ C) .

Another, longer but clearer, style of presenting equivalence proofs consists in writing the sentences on separate lines, indicating in the margin the grounds for the step. Our last chain becomes:

1. ¬[(A ∧ B) ∨ ¬C]

2. ¬(A ∧ B) ∧ ¬¬C De Morgan’s law for disjunction,

3. ¬(A ∧ B) ∧ C double negation law,

4. (¬A ∨ ¬B) ∧ C De Morgan’s law for conjunction,

5. (¬A ∧ C) ∨ (¬B ∧ C) right distributive law for conjunction.

Often, several simple steps are combined into one. We can, for example, drop double negations immediately without special indication; we can accordingly pass directly from 1. to 3. in the proof above. We also need not use separate steps for changing (via associativity) the grouping in repeated conjunctions and disjunctions; in fact, these groupings can be ignored (cf. 2.3.0 “Repeated Conjunctions and Disjunction”). Similarly, we need not bother with explicit changes of order or (via commutativity) in conjunctions and disjunctions, or with explicit deletions of repeated conjuncts, or disjuncts (via idempotence). Once you get the hang of it you can assimilate these steps into others.

Pushing-In Negations

We mentioned already the pushing-in of negations, via De Morgan’s laws:

From ¬(A ∧ B) to ¬A ∨ ¬B . From ¬(A ∨ B) to ¬A ∧ ¬B .

This replaces one occurrence of ‘¬’, whose scope is A ∧ B or A ∨ B, by two occurrence with smaller scopes: A and B. In the last example. negation is pushed inside in the passage from 1 to 2; then it is pushed again in the passage from 3 to 4.

As long as there are components of the form ¬(A ∧ B), or ¬(A ∨ B), we can push negation in. By repeating this process, we will eventually get a sentence that has no components of these forms. (The mathematical proof of this claim will not be given here. But intuitively it is very clear, especially after having worked out a few examples.) In addition, we can always drop double negations. At the end the process there will be no component with a string of more than one negation. Consequently, any sentential expression whose connectives are among ‘¬’, ‘∧’, and ‘∨’, can be transformed into an equivalent one, in which negation applies only to the sentential variables.

If several negation-occurrences can be pushed in, we can choose any of them. The final outcome does not depend on the choices of the pushed-in negations, provided that negation is pushed all the way in, all double negations are dropped and no other laws are applied. But the number of steps can vary. It is advisable to start with the outermost negations; because as the negation moves in it will form with the inner ones double negations, which can be immediately dropped. Here is an example to illustrate all these points. Given ¬[A ∧ ¬(B ∧ C)], we can start by pushing-in the inner (i.e., second) negation:

¬[A ∧ ¬(B ∧ C)], ¬[A ∧ (¬B ∨ ¬C)], ¬A ∨ ¬(¬B ∨ ¬C), ¬A ∨ (B ∧ C) If we start by pushing-in the outermost negation, we get:

¬[A ∧ ¬(B ∧ C)], ¬A ∨ ¬¬(B ∧ C), ¬A ∨ (B ∧ C)

De Morgan’s Laws for Repeated Conjunctions and Disjunctions The law for conjunctions with more than two conjuncts is:

¬(A¹∧ A²∧ . . . ∧ Aⁿ) ≡ ¬A¹∨ ¬A²∨ . . . ∨ ¬Aⁿ

Here we are ignoring the grouping, since we are interested in logical equivalence only. The equivalence can be deduced by repeated applications of the two-conjunct version of De Mor-gan. For example, if n = 3, we have:

¬[(A¹∧ A²)∧ A³] ≡ ¬(A¹∧ A²)∨ ¬A³ ≡ (¬A¹∨ ¬A²)∨ ¬A³ It is also easily derivable by considering truth-values:

The left-hand side is true iff A1∧ A²∧ . . . ∧ Aⁿ is false, i.e., iff at least one of the Ai’s is false. The right-hand side is true iff at least one of the sentences

¬Aⁱ is true, i.e., iff at least one of the Ai’s is false. Hence, the left-hand side is true iff the right-hand side is.

The case of disjunctions is the exact dual:

¬(A¹∨ A²∨ . . . ∨ Aⁿ) ≡ ¬A¹∧ ¬A²∧ . . . ∧ ¬Aⁿ

Duality applies throughout; each of the above observations or claims has a dual.

Pushing-In Conjunctions

Conjunctions are pushed-in by distributing repeatedly conjunction over disjunction. As ex-plained already, we do it by going left-to-right in the distributive laws for conjunction:

A∧ (B ∨ C) ≡ (A ∧ B) ∨ (A ∧ C) (B∨ C) ∧ A ≡ (B ∧ A) ∨ (C ∧ A)

It increases the size (both ‘A’ and ‘∧’ occur once on the left, but twice on the right), but decreases the scope of the conjunction (instead of B∨ C we have the scopes B and C). It also increases the scope of the disjunction. Here is an example of pushing-in conjunctions:

A∧ [(¬B ∨ C) ∧ ¬D] ≡ A ∧ [(¬B ∧ ¬D) ∨ (C ∧ ¬D)] ≡ (A ∧ ¬B ∧ ¬D) ∨ (A ∧ C ∧ ¬D) If we keep pushing-in conjunctions, we must eventually arrive at a point where no further pushing-in is possible; at that stage, there are no components either of the form A∧ (B ∨ C) or of the form (B∨ C) ∧ A. You may convince yourself that the process must terminate by working out a few examples; the formal proof, which will not be given here, is far from trivial.

WATCH IT: Distributing the conjunction in ¬[(B ∨ C) ∧ A] yields ¬[B∧A ∨ C ∧A]. But you cannot distribute the conjunction in ¬(B ∨ C) ∧ A , because the first conjunct is not a

disjunction, but a negation. You can however simplify by pushing-in the negation:

¬B ∧ ¬C ∧ A .

Simultaneous Distributing of Conjunctions: Repeated applications of the distributive laws to

(A1∨ A²)∧ (B¹∨ B²) yields the chain:

(A1∨A²)∧(B¹∨B²)≡ (A¹∧(B¹∨B²))∨(A²∧(B¹∨B²))≡ (A¹∧B¹)∨(A¹∧B²)∨(A²∧B¹)∨(A²∧B²) . The final sentence is a disjunction of all the conjunctions Ai∧B^j. This generalizes to arbitrary disjunctions:

(A1∨ A²∨ . . . ∨ A^m)∧ (B¹∨ B²∨ . . . ∨ Bⁿ)

is logically equivalent to the disjunction of all the conjunctions in which one of the Ais is

“conjuncted” with one of the Bjs:

(A1∧B¹)∨ . . . ∨ (A¹∧Bⁿ)∨ (A²∧B¹)∨ . . . ∨ (A²∧Bⁿ)∨ . . . ∨ (A^m∧B¹)∨ . . . ∨ (A^m∧Bⁿ) Altogether there are m·n disjuncts. It generalized further to more than two conjuncts:

(A1∨ . . . ∨ A^m)∧ (B¹∨ . . . ∨ Bⁿ)∧ (C¹∨ . . . ∨ C^p) is logically equivalent to the disjunctions of all conjunctions of the form

Ai∧ B^j∧ C^k ,

where 1 ≤ i ≤ m, 1 ≤ j ≤ n, 1 ≤ k ≤ p ; here we get m·n·p disjuncts (there are m choices for i, n choices for j, and p choices for k). In particular,

(A1∨A²)∧ (B¹∨B²)∧ (C¹∨C²) is logically equivalent to a disjunctions of 8 conjunctions:

A1∧B¹∧C¹ ∨ A¹∧B¹∧C² ∨ A¹∧B²∧C¹ ∨ A¹∧B²∧C²

∨ A²∧B¹∧C¹ ∨ A²∧B¹∧C² ∨ A²∧B²∧C¹ ∨ A²∧B²∧C²

Pushing-In Disjunctions

Pushing-in disjunctions, which is carried out via the distributive laws for disjunction, is the exact dual. It reduces the scopes of ∨, but enlarges those of ∧. For example, two steps of pushing-in disjunction yield:

A∨ [(¬B ∧ C) ∨ ¬D] ≡ A ∨ [(¬B ∨ ¬D) ∧ (C ∨ ¬D)] ≡ (A ∨ ¬B ∨ ¬D) ∧ (A ∨ C ∨ ¬D)

Repeated pushing-in of disjunction terminates at a stage where there are no components either of the form A∨ (B ∧ C) or of the form (B ∧ C) ∨ A.

As in the case of conjunction, we can distribute, at one go, disjunction over several conjunc-tions:

(A1∧ A²∧ . . . ∧ A^m)∨ (B¹ ∧ B²∧ . . . ∧ Bⁿ) is logically equivalent to:

(A1∨B¹)∧ . . . ∧ (A¹∨Bⁿ)∧ (A²∨B¹)∧ . . . ∧ (A²∨Bⁿ)∧ . . . ∧ (A^m∨B¹)∧ . . . ∧ (A^m∨Bⁿ) And this generalizes further to more than two disjuncts.

Homework

2.14 Let G be the sentence:

¬[(A ∨ ¬B) ∧ ¬C] ∧ ¬[¬C ∨ (¬D ∧ E)]

Construct the following sentences:

G1: Obtained from G by pushing-in negations all the way.

G2: Obtained from G1 by pushing-in conjunctions all the way.

G3: Obtained from G1 by pushing-in disjunctions all the way.

If possible, simplify G2 and G3 by using the idempotence laws.

2.15 Let H = ¬G, where G is as in 2.15 above. Construct H¹ –obtained from H by pushing-in negations all the way, H2 –obtained from H1 by pushing-in conjunctions all the way, and H3 obtained from H1 by pushing-in disjunctions all the way.

WATCH IT: It is inadvisable to mix pushing-in conjunctions and pushing-in disjunctions.

The first reduces the scopes of ∧ and enlarges those of ∨. The second has the opposite effect. And both increase sentence size. If you interlace them you will get longer and longer sentences. For example:

A∧ (B ∨ C) ≡ (A ∧ B) ∨ (A ∧ C) ≡ [(A ∧ B) ∨ A] ∧ [(A ∧ B) ∨ C] ≡ {[(A ∧ B) ∨ A] ∧ (A ∧ B)} ∨ {[(A ∧ B) ∨ A] ∧ C} ≡ . . .

Occasionally you may want to apply the two kinds of pushing-in to separate components, or to combine them with other operations in between.

Pulling Out Negations and Common Factors

By applying De Morgan’s laws from right to left, we can pull out negations: ¬A ∨ ¬B is converted to ¬(A ∧ B), and ¬A ∧ ¬B is converted to ¬(A ∨ B). This generalizes to disjunctions of more than two disjuncts; similarly for conjunctions; i.e., we can replace

¬A¹∨ ¬A²∨ . . . ∨ ¬Aⁿ by ¬(A¹∧ A²∧ . . . ∧ Aⁿ) ,

¬A¹∧ ¬A²∧ . . . ∧ ¬Aⁿ by ¬(A¹∨ A²∨ . . . ∨ Aⁿ) .

Similarly, by applying the distributive laws in the right-to-left direction we pull out a common factor; this is a common conjunct–if the law is for conjunction, a common disjunct–if the law is for disjunction. Thus we can replace:

(A∧B) ∨ (A∧C) by A∧ (B ∨ C) ,

(A∨B) ∧ (A∨C) by A∨ (B ∧ C) .

It generalizes to more than two disjuncts that share a common conjunct, and to more than two conjuncts that share a common disjunct:

(A∧B) ∨ (A∧C) ∨ (A∧D) is replaceable by A∧ (B ∨ C ∨ D) . (A∨B) ∧ (A∨C) ∧ (A∨D) is replaceable by A∨ (B ∧ C ∧ D) .

Since each pull-out step reduces the size of the sentence, repeated pull-outs–whether of common conjuncts or of common disjuncts, or of both–must terminate at a stage where no further pulling out is possible. In the case of more than two disjuncts (or more than two conjuncts), the final outcome of repeated pull-out can be highly sensitive to the order in which it is done. Consider for example:

(A∧B) ∨ (A∧C) ∨ (D∧C) .

If we group together the first two disjuncts and pull out A, we get:

[A∧(B∨C)] ∨ (D∧C)

and no further pulling out is possible. But if we group together the second and the third disjunct and pull out C, we get:

(A∧B) ∨ [(A∨D)∧C]

and, again, no further pulling out is possible. The two final outcomes are quite different; they are of course logically equivalent, but the equivalence is not obvious at first glance.