Reasoning

To be able to reason, we need to find the rules which will allow us to eliminate ambiguity. That’s what principles of logic are designed to do and that’s the subject of this chapter.

Table of Contents

Predicates

The simplest strategy to eliminate confusion is to avoid statements having multiple meanings. In other words, there should be no space for interpretation, everyone’s understanding of the given statement should be exactly the same.

To simplify things further, let’s consider statements that can be only true or false.

From now on, the statements having these two features will be called predicates.

Now you may wonder why true and false only. Reasoning simply means establishing what relations exist between statements. If statements can only be true or false, doing that is much simpler.

Implication

Let’s take a look at an example of relation between predicates. Imagine we have a bowl full of balls that can only be black or white, and we define two predicates:

P := all balls in the bowl are black

Q := a ball randomly chosen from the bowl is black

P and Q are simply shorthands for the respective predicate. If predicate is true, we will indicate that with number 1. False will be represented by zero e.g. P=1, Q=0.

Let’s examine all the possibilities:

(P=0) and (Q=0)
This is possible because if there are some white balls in the bowl (P=0) then a randomly chosen one may be white (Q=0).
(P=0) and (Q=1)
This is possible because if the bowl contains some black and some white balls (P=0), picking up a black one is in the cards (Q=1).
(P=1) and (Q=0)
All balls in the bowl are black (P=1) and a randomly chosen one is white (Q=0). That absolutely cannot happen.
(P=1) and (Q=1)
All the balls in the bowl are black (P=1) and a randomly chosen ball is black (Q=1). This is certain.

To summarize these results we will define yet another predicate:

R(x,y) := “(P=x) and (Q=y) is possible”

where x and y are numerical variables which can be only 1 or 0. With that in mind we can put the results in the following table:

P := ”all balls in the bowl are black”	Q := ”a ball randomly chosen from the bowl is black”	R(x,y) := “(P=x) and (Q=y) is possible”
0	0	1
0	1	1
1	0	0
1	1	1

The table describes a relation between P and Q defined by the behavior of the predicate R.

How can we tell if the relation between some other arbitrary predicates A and B is the same as the relation between P and Q?

We cannot use “(A=x) and (B=y) is possible” because its behavior depends on what A and B actually mean and we assumed A and B are not specified. We have to get rid of this dependency.

Let’s introduce an implication operator $\boldsymbol{\hat{\Longrightarrow}}$ which turns A and B into a new predicate. By definition A $\boldsymbol{\hat{\Longrightarrow}}$ B will always follow the same pattern as R irrespective of what A and B mean:

A	B	A $\boldsymbol{\hat{\Longrightarrow}}$ B
0	0	1
0	1	1
1	0	0
1	1	1

Note that if we insert P in place of A and Q in place of B then we can write:

(P $\boldsymbol{\hat{\Longrightarrow}}$ Q)=1

because we know that combination P=1 and Q=0 (the third row in the table above) is not possible and only then P $\boldsymbol{\hat{\Longrightarrow}}$ Q would be 0.

Imagine that for some other concrete predicates R and S we have (R $\boldsymbol{\hat{\Longrightarrow}}$ S)=1. It is clear that the relation between R and S is of the same nature as between P and Q.

From now on, this particlular relation will be called implication.

To shorten the notation we will introduce another symbol: $\boldsymbol{\Rightarrow}$ and define
“R $\boldsymbol{\Rightarrow}$ S” as the equivalent of: “(R $\boldsymbol{\hat{\Longrightarrow}}$ S)=1″.

In plain English R $\boldsymbol{\Rightarrow}$ S means: if R is true then S must be true as well.

Note that we don’t have to say R $\boldsymbol{\Rightarrow}$ S is true, because “is true” is already built into the statement. We just say: R $\boldsymbol{\Rightarrow}$ S.

Finally, note that to prove that the implication A $\boldsymbol{\Rightarrow}$ B holds, it is enough to assume that A=1 and check whether B=1 follows. If it does, then A implies B and checking what happens for A=0 is redundant. Why?

Because irrespective whether we have:
(A=0) and (B=0) must coincide
or
(A=0, B=0) and (A=0, B=1) are both possible,
the statement below follows:
(A $\boldsymbol{\hat{\Longrightarrow}}$ B)=1

Basic logical operators

Sometimes we want to know if several statements are true at the same time, and sometimes we are happy with at least one of them to be true. For these purposes we use operators of conjunction and disjunction.

Conjunction

Let’s start with an example:

P := the object is black

Q := the object is a ball

E := P and Q

E is true only when the object is a black ball. Instead of the English word “and” mathematicians use the symbol $\boldsymbol{\wedge}$ .

For some unspecified predicates A and B, the results of A $\boldsymbol{\wedge}$ B can be defined using the following table:

A	B	A $\boldsymbol{\wedge}$ B
0	0	0
0	1	0
1	0	0
1	1	1

After adding a third predicate C, we have:

A $\boldsymbol{\wedge}$ B	C	(A $\boldsymbol{\wedge}$ B) $\boldsymbol{\wedge}$ C
0	0	0
0	1	0
1	0	0
1	1	1

In this case we can see that if we have n predicates connected with $\boldsymbol{\wedge}$ , all of them must be true for the whole statement to be true. It is clear then that it doesn’t matter in what order we apply $\boldsymbol{\wedge}$ :

A $\boldsymbol{\wedge}$ B $\boldsymbol{\wedge}$ C = C $\boldsymbol{\wedge}$ B $\boldsymbol{\wedge}$ A = B $\boldsymbol{\wedge}$ A $\boldsymbol{\wedge}$ C = etc.

Disjunction

Let’s use the same predicates P and Q as in the example above and define a new statement:

D := P or/and Q

D is true when either the object is black or it is a ball, or both. From now on we will use the symbol $\boldsymbol{\vee}$ instead of “or/and”. Mathematics is universal after all.

We can summarize the behavior of disjunction in a table. A and B are some unspecified predicates and the column on the right defines “A $\boldsymbol{\vee}$ B”:

A	B	A $\boldsymbol{\vee}$ B
0	0	0
0	1	1
1	0	1
1	1	1

After we add another unspecified predicate C to the mix, it is easy to check that:

A $\boldsymbol{\vee}$ B	C	(A $\boldsymbol{\vee}$ B) $\boldsymbol{\vee}$ C
0	0	0
0	1	1
1	0	1
1	1	1

Looking at this table, it is clear that if we have n predicates connected with $\boldsymbol{\vee}$ , then the resulting statement will be true when at least one of them is true. From this it follows that it doesn’t matter in what order we apply $\boldsymbol{\vee}$ :

A $\boldsymbol{\vee}$ B $\boldsymbol{\vee}$ C = C $\boldsymbol{\vee}$ B $\boldsymbol{\vee}$ A = B $\boldsymbol{\vee}$ A $\boldsymbol{\vee}$ C = etc.

Negation

Apart from combining predicates, sometimes it is useful to negate one. This operation is represented by the symbol ~ and can be summarized using the following table:

A	~A
0	1
1	0

Note that two negations cancel each other out:
~(~A)=A

Equivalence

Let’s take another look at the following example:

P := all balls in the bowl are black

Q := a ball randomly chosen from the bowl is black

We have P $\boldsymbol{\Rightarrow}$ Q, however, it doesn’t work the other way around Q $\boldsymbol{\not \Rightarrow}$ P because if a randomly chosen ball is black, it doesn’t mean that all balls in the bowl are necessarily black. For these specific predicates the implication doesn’t work in both directions.

Let’s take a look at another example:

P’ := all balls in the bowl are black

Q’ := a ball randomly chosen from the bowl is always black

In this case, implication works both ways, so we have:
(P’ $\boldsymbol{\Rightarrow}$ Q’) $\boldsymbol{\wedge}$ (Q’ $\boldsymbol{\Rightarrow}$ P’)

How can we tell if the relation between some arbitrary A and B is the same as for P’ and Q’?

As we know from the section on implication:
(P’ $\boldsymbol{\Rightarrow}$ Q’) $\boldsymbol{\wedge}$ (Q’ $\boldsymbol{\Rightarrow}$ P’)

is equivalent to:
(P’ $\boldsymbol{\hat{\Longrightarrow}}$ Q’) =1 $\boldsymbol{\wedge}$ (Q’ $\boldsymbol{\hat{\Longrightarrow}}$ P’) = 1

That suggests we should take a closer look at a predicate:
(A $\boldsymbol{\hat{\Longrightarrow}}$ B) $\boldsymbol{\wedge}$ (A $\boldsymbol{\hat{\Longrightarrow}}$ B)
for arbitrary A and B.

Using our knowledge about operators of implication ( $\boldsymbol{\hat{\Longrightarrow}}$ ) and conjunction ( $\boldsymbol{\wedge}$ ) we can put together the following table:

A	B	A $\boldsymbol{\hat{\Longrightarrow}}$ B	B $\boldsymbol{\hat{\Longrightarrow}}$ A	(A $\boldsymbol{\hat{\Longrightarrow}}$ B) $\boldsymbol{\wedge}$ (B $\boldsymbol{\hat{\Longrightarrow}}$ A)
0	0	1	1	1
0	1	1	0	0
1	0	0	1	0
1	1	1	1	1

Concrete predicates P’ and Q’ must be either both true or both false, so in the language of the table above we have:

(P’ $\boldsymbol{\hat{\Longrightarrow}}$ Q’) $\boldsymbol{\wedge}$ (Q’ $\boldsymbol{\hat{\Longrightarrow}}$ P’) = 1

In other words, the second and third row cannot happen. We will call a relation like that equivalence and use the symbol $\boldsymbol{\Leftrightarrow}$ to express it.

From now on, the predicate:
A $\boldsymbol{\Leftrightarrow}$ B

by definition will be identical to:
(A $\boldsymbol{\hat{\Longrightarrow}}$ B) $\boldsymbol{\wedge}$ (B $\boldsymbol{\hat{\Longrightarrow}}$ A) = 1

Circular logic

Perhaps you heard somewhere that circular logic is BAD! Yet it seems we happily use it!

Example:

P := all balls in the bowl are black

Q := a ball randomly chosen from the bowl is always black

To prove the equivalence between P and Q, we simply assume that P is true and check if Q follows and then we assume Q to be true and check if P follows. In the process we make assumptions without knowing whether P and/or Q are actually true or not.

Is circular logic bad then, or not?

The procedure described is valid for establishing the equivalence. After proving that P is equivalent to Q, we know that P and Q must be either both true or both false.

Claiming that both are true is a fallacy because equally well both may be false.

Reductio ad absurdum

The implications we have seen so far were obvious, yet sometimes it is not that straightforward to see if the implication holds. Thankfully, it is possible to approach the problem indirectly.

Using knowledge about operators of implication ( $\boldsymbol{\hat{\Longrightarrow}}$ ), negation (~) and conjunction ( $\boldsymbol{\wedge}$ ) we can put together the following table:

A	B	A $\boldsymbol{\hat{\Longrightarrow}}$ B	~ (A $\boldsymbol{\hat{\Longrightarrow}}$ B)	A $\boldsymbol{\wedge}$ ~B
0	0	1	0	0
0	1	1	0	0
1	0	0	1	1
1	1	1	0	0

It is clear that we always have:
~(A $\boldsymbol{\hat{\Longrightarrow}}$ B) $\boldsymbol{\Leftrightarrow}$ (A $\boldsymbol{\wedge}$ ~B)

In other words, predicates ~(A $\boldsymbol{\hat{\Longrightarrow}}$ B) and (A $\boldsymbol{\wedge}$ ~B) are either both true or both false.

Let’s apply this relation to an example:

P := all balls in the bowl are black

Q := a ball randomly chosen from the bowl is black

Is it possible that all balls are black (P=1), yet a randomly chosen one is not (~Q=1)?

No, it is not possible:
P $\boldsymbol{\wedge}$ ~Q=0

and we know that the statement above is equivalent to:
~(P $\boldsymbol{\hat{\Longrightarrow}}$ Q)=0

so we must have:
(P $\boldsymbol{\hat{\Longrightarrow}}$ Q)=1

Our example is extremely simple, and we don’t have to resort to reductio ad absurdum to prove the implication. But as we will see later on, in some situations it is very convenient.

Summary

The second opening quotation from the chapter “Models of reality” suggests that counting on fingers is a good starting point for building models, yet we started with reasoning instead. We had to, otherwise counting fingers wouldn’t take us anywhere.

It looks like the anonymous sage was wrong after all. This sort of thing is going to happen many times moving forward. An assumption which sounds reasonable at a time, will turn out either completely wrong or inaccurate.

Now that we know the basic rules of reasoning, we are ready to take a closer look at counting.