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propositional equivalences

Module by: Ian Barland, John Greiner. E-mail the authors

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Summary: How to use identities to determine whether two propositional formulas are equivalent.

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Propositional Equivalences

What are the roots of x3−4x x 3 4 x ? Well, in high-school algebra you learned how to deal with such numeric formulas:

x3−4x x 3 4 x (1)
= xx2−4 [factor out x] = x x 2 4 [factor out x] (2)
= xx−2x+2 [identity a2−b2=a+ba−b , with a being x , and b being 2 .] = x x 2 x 2 [identity = a 2 b 2 a b a b , with abeingx, andbbeing2.] (3)
(This last expression happens to be useful since it is in a form which lets us read off the roots 0, +2, -2.) The rules of algebra tell us that these three different formulas are all equivalent. We are distinguishing between syntax (the expression itself, as data), and semantics (what the expression means). Usually, when presented with syntax, one is supposed to bypass that and focus on its meaning (e.g., reading a textbook). However, in logic and postmodern literature alike, we are actually studying the interplay between syntax and semantics. The general gist is that in each step, you rewrite subparts of your formula according to certain rules ("replacing equals with equals").

Well, we can use a similar set of rules about rewriting formulas with equivalent ones, to answer the questions of whether two formulas are equal, or whether a formula is a tautology. George Boole was the first to realize that true and false are just values in the way that numbers are, and he first codified the rules for manipulating them; thus Boolean algebra is named in his honor.

Aside:

The " algebra" comes from true, false, ∧∧, ∨∨ having some very specific properties similar to those of numbers, ×, +.

Figure 1: George Boole (1815-1864)
Figure 1 (boole.png)

Again, each individual step consists of rewriting a formula according to certain rules. So, just what are the rules for manipulating Boolean values? We'll start with an example.

Example 1

Table 1
1a ∧ false ∨ b ∧ truea ∧ ∨ b ∧  
2≡false ∨ b ∧ true≡ ∨ b ∧ Dominance of false over ∧∧
3≡b ∧ true ∨ false≡b ∧ ∨ Commutativity of ∨∨
4≡b ∧ true≡b ∧ Identity element for ∨∨ is false
5≡b≡bIdentity element for ∧∧ is true

Thus we have a series of equivalent formulas, with each step justified by citing a propositional equivalence . By and large, the equivalences are rather mundane. A couple are surprisingly handy; take a moment to consider DeMorgan's laws.

Table 2
¬φ ∧ ψ ≡ ¬φ ∨ ¬ψ¬φ ∧ ψ≡¬φ ∨ ¬ψ ¬φ ∨ ψ ≡ ¬φ ∧ ¬ψ¬φ ∨ ψ≡¬φ ∧ ¬ψ

(Try φφ being "Leprechauns are green", and ψψ being "Morgana Le Fay likes gold". Do these laws make sense, for each of the four possible T/F cases?) Augustus DeMorgan was also an important figure in the formalization of logic.

Figure 2: Augustus DeMorgan (1806-1871)
Figure 2 (demorgan.jpg)

Here are two more examples. The first is a proof of one of the laws from the given list, using others from the list.

Example 2

Absorption of ∨∨

Table 3
1a ∧ b ∨ ba ∧ b ∨ b 
2≡a ∧ b ∨ b ∧ true≡a ∧ b ∨ b ∧ Identity of ∧∧
3≡b ∧ a ∨ b ∧ true≡b ∧ a ∨ b ∧ Commutativity of ∨∨
4≡b ∧ a ∨ true≡b ∧ a ∨ Distributivity of ∧∧ over ∨∨
5≡b ∧ true≡b ∧ Dominance of ∨∨
6≡b≡bIdentity of ∧∧

Example 3

Contrapositive

Table 4
1a → ba → b 
2≡¬a ∨ b≡¬a ∨ bDefinition of →→
3≡b ∨ ¬a≡b ∨ ¬aCommutativity of ∨∨
4≡¬¬b ∨ ¬a≡¬¬b ∨ ¬aDouble Complementation
5≡¬b → ¬a≡¬b → ¬aDefinition of →→

Exercise 1

Show that the "Absorption of ∧∧" equivalence holds, given the other equivalences. I.e., show a ∨ b ∧ b ≡ ba ∨ b ∧ b≡b.

Solution

Table 5
1a ∨ b ∧ ba ∨ b ∧ b 
2≡a ∨ b ∧ b ∨ false≡a ∨ b ∧ b ∨ Identity of ∨∨
3≡b ∨ a ∧ b ∨ false≡b ∨ a ∧ b ∨ Commutativity of ∨∨
4≡b ∨ a ∧ false≡b ∨ a ∧ Distributivity of ∨∨ over ∧∧
5≡b ∨ false≡b ∨ Dominance of ∧∧
6≡b≡bIdentity of ∨∨

Compared to proofs using truth tables, Boolean algebra gives us much shorter proofs. But, determining which equivalence to use in the next step of a proof can be difficult. In this case, compare the solution for this exercise to the previous absorption proof. These two proofs have a special dual relationship described in the next section.

Exercise 2

Show that the modus ponens rule, a ∧ a → b → ba ∧ a → b → b always holds. I.e., show that it is a tautology, and thus equivalent to true.

Solution

Table 6
1a ∧ a → b → ba ∧ a → b → b 
2≡a ∧ ¬a ∨ b → b≡a ∧ ¬a ∨ b → bDefinition of →→
3≡a ∧ ¬a ∨ a ∧ b → b≡a ∧ ¬a ∨ a ∧ b → bDistributivity of ∨∨ over ∧∧
4≡false ∨ a ∧ b → b≡ ∨ a ∧ b → bComplement
5≡a ∧ b ∨ false → b≡a ∧ b ∨ → bCommutativity of ∨∨
6≡a ∧ b → b≡a ∧ b → bIdentity of ∨∨
7≡¬ a ∧ b ∨ b≡¬ a ∧ b ∨ bDefinition of →
8≡¬a ∨ ¬b ∨ b≡¬a ∨ ¬b ∨ bDeMorgan's law
9≡¬a ∨ ¬b ∨ b≡¬a ∨ ¬b ∨ bAssociativity of ∨∨
10≡¬a ∨ b ∨ ¬b≡¬a ∨ b ∨ ¬bCommutativity of ∨∨
11≡¬a ∨ true≡¬a ∨ Complement
12≡true≡Dominance of ∨∨

So, what would it mean to use Boolean algebra as reasoning for WaterWorld? That is, if you wanted to show that G-safeG-safe was true, how would you do that using Boolean algebra? It would mean starting with a big rule involving the conjunction of all the WaterWorld domain axioms (ρρ), and the board's observed state (ψψ), and showing that these were equivalent to G-safeG-safe (… and the domain axioms and state). That is, ρ ∧ ψ ≡ G-safe ∧ ρ ∧ ψρ ∧ ψ≡G-safe ∧ ρ ∧ ψ.

Duals (optional)

Duals: a symmetry between ∧∧, ∨∨ mediated by ¬¬.

Looking at the provided propositional equivalences, you should notice a strong similarity between those for ∨∨ and those for ∧∧. Take any equivalence, swap ∨∨s and ∧∧s, swap true and falses, and you'll have another equivalence! For instance, there are two flavors of DeMorgan's law, which are just duals of each other:

Table 7
¬φ ∧ ψ ≡ ¬φ ∨ ¬ψ¬φ ∧ ψ≡¬φ ∨ ¬ψ ¬φ ∨ ψ ≡ ¬φ ∧ ¬ψ¬φ ∨ ψ≡¬φ ∧ ¬ψ

Aside:

In terms of circuit diagrams, we can change each AND gate to an OR gate and add negation-bubbles to each gate's inputs and outputs. The principle of duality asserts that this operation yields an equivalent circuit.

The idea of duality is more general than this. For example, polyhedra have a natural dual of interchanging the role of vertices and faces.

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