# Inductively Strong Arguments

Recall that an argument $X\Rightarrow Y$ is valid when there is no truth value assignment that makes $X$ true and $Y$ false. There are different ways to express that an argument is valid. The following three statements are equivalent (meaning that 1 is true if, and only if, 2 is true; 2 is true if and only if 3 is true; and 3 is true if, and only if, 1 is true):

1. The argument with premise $X$ and conclusion $Y$ is valid (denoted $\varphi\models\psi$)
2. $X\rightarrow Y$ is a tautology.
3. $X\wedge\neg Y$ is a contradiction.

In this section, we are interested in arguments that are inductively strong. This is weaker than being valid. For instance, the following argument is not valid:

Ann brought her laptop to the first 10 lectures. Therefore, Ann will bring her laptop to the next lecture.

However, the above argument is a strong argument.

The first thing to note is that whether an argument is inductively strong will depend on the stochastic truth table. This is very different than when evaluating whether an argument is valid. To determine validity of an argument $X\Rightarrow Y$, you need to construct a truth table with columns for $X$ and $Y$. This truth table is completely determined by the formulas $X$ and $Y$. Evaluating whether $X\Rightarrow Y$ is inductively strong requires more information than the truth table with columns for $X$ and $Y$. An argument may be inductively strong according to one way of assigning probabilities to the rows of its truth table, but not inductively strong according to a different way of assigning probabilities to the rows of its truth table.

The first idea that one might have to define inductively strong arguments is: $X\Rightarrow Y$ is inductively strong when $X\rightarrow Y$ is probable. To say that a formula is "probable" means that the probability of the formula is above some threshold. For instance, the probability of the formula is greater than $0.5$. However, this definition does not capture the notion of inductive strength that we are after. The following well-known examples from Skyrms, Choice and Chance, Chapter II illustrate the problem with classifying $P\Rightarrow C$ as inductively strong when the $P\rightarrow C$ is probable. First of all, recall that $P\rightarrow C$ is tautologically equivalent to $\neg P\vee C$. So, $P\rightarrow C$ is probable if, and only if, $\neg P\vee C$ is probable.

Example

Let $P$ be "there is a man in Cleveland that is 2,000 years old", and $C$ be "there is a man in Cleveland with three heads that is 2,000 years old". Intuitively, the argument $P\Rightarrow C$ is not inductively strong. However, the disjunction $\neg P\vee C$ is probable. Since $P$ is very unlikely likely to be true, $\neg P$ is very likely to be true, so $\neg P\vee C$ is very probable. So, the above definition would incorrectly classify this as an inductively strong argument.

Example

Let $P$ be "there is a man in Cleveland that is 1,999 years and 11 months old and in good health", and $C$ be "there will never be a man in Cleveland that is 2,000 years old". Intuitively, the argument $P\Rightarrow C$ is not inductively strong. However, the conjunction $\neg P\vee C$ is probable. Since $\neg C$ is very likely to be true, $\neg P\vee C$ is also very likely to be true. So, the above definition would incorrectly classify this as an inductively strong argument.

The crucial problem in both examples that $\neg P\vee C$ may be probable (and so $P\rightarrow C$ is probable) even though there is no evidential relationship between $P$ and $C$. This suggests the following definition of inductive strength: $C$ is probable given that $P$ is true (or under the supposition that $P$ is true). That is, the argument with premise $P$ and conclusion $C$ is inductively strong when $Pr(C\mid P)$ is probable. This definition matches the intuitive judgements in the above two examples: In both of the above examples, $Pr(C\mid P)$ is low (e.g., the probability that there is 2,000 year old man with three heads is low under the supposition that there is a 2,000 year old man in Cleveland). This suggests the first component of an inductively strong argument:

Evidential Support

Suppose that $X$ and $Y$ are formulas and $t$ is a number between $0.5$ and $1$. We say that $X$ evidentially supports $Y$ to degree $t$ in a stochastic truth table if $Pr(Y\mid X) > t$. When $t=0.5$, we say that $Pr(Y\mid X)$ is probable and that $X$ evidentially supports $Y$.

Evidential support is not the only property that inductively strong arguments satisfy. To illustrate, consider the following argument:

1. Let the premise $P$ be "Bob (who is male) is taking a new birth control pill."
2. Let the conclusion $C$ be "Bob (who is male) does not get pregnant."

Then, $Pr(C\mid P)=1$, so according to the above definition $P$ evidentially supports $C$. However, $P\Rightarrow C$ is not an inductively strong argument. As in the above two examples, there is no evidential relationship between $P$ and $C$. In this case, the problem is that $Pr(C)=1=Pr(C\mid P)$. That is, learning that $P$ is true does not change the probability that $C$ is true. In such a case we say that $C$ and $P$ are independent.

Independent Formulas

Given a stochastic truth table, we say that $X$ and $Y$ are independent when $Pr(X\mid Y)=Pr(X)$.

There are alternative characterizations of independence:

Alternative Characterizations of Independence

Given a stochastic truth table, for all formulas $\varphi$ and $\psi$, the following are equivalent:

1. $X$ and $Y$ are independent ($Pr(X\mid Y)=Pr(X)$).
2. $Pr(Y\mid X)=Pr(Y)$.
3. $Pr(X\wedge Y)=Pr(X)Pr(Y)$.

The second aspect of an inductively strong argument is that the premises are relevant to the conclusion:

Positive/Negative Relevance

Given a stochastic truth table, we say that:

1. $X$ is positively relevant to $Y$ when $Pr(Y\mid X) > Pr(Y)$; and
2. $X$ is negatively relevant to $Y$ when $Pr(Y\mid X) < Pr(Y)$.

Putting everything together, the argument $X\Rightarrow Y$ is inductively strong when:

1. $X$ evidentially supports $Y$ (i.e., $Pr(Y\mid X)>0.5$).
2. $X$ is positively relevant to $Y$ (i.e., $Pr(Y\mid X)>Pr(Y)$).
3. $X\rightarrow Y$ is not valid.

The reason that we do not define valid arguments as inductively strong is because an argument can be valid because its premises are contradictory. For instance, $(P\wedge\neg P)\models Q$ is valid. However, since $Pr(P\wedge\neg P)=0$, $Pr(Q\mid P\wedge\neg P)$ is undefined, so $P\wedge \neg P$ does not evidentially support $Q$ and $P\wedge\neg P$ is not positively relevant to $Q$.

## Practice Questions​

In the stochastic truth table below, you can change the probabilities assigned to each row. Recall that the sum of the numbers assigned to each row must be 1 and each number must be greater than or equal to 0. The probabilities of the (conditional) formulas are updated when the stochastic truth table changes. When you hover over a formula with your mouse, the rows of the truth table where that formula is true is highlighted. Use this interactive tool to solve the following problems:

1. Find a stochastic truth table such that $P$ and $Q$ are independent.
2. Find a stochastic truth table such that $Q$ is positively relevant to $P$.
3. Find a stochastic truth table such that $Q$ is negatively relevant to $P$.
4. Find a stochastic truth table such that $P\vee Q$ evidentially supports $P$, but $P\vee Q$ does not evidentially supports $Q$.
$P$$Q$
$\mathsf{T}$$\mathsf{T}$
$\mathsf{T}$$\mathsf{F}$
$\mathsf{F}$$\mathsf{T}$
$\mathsf{F}$$\mathsf{F}$
The sum of the numbers assigned to each row must be 1.
$Pr(P)$ $\ =\$ undefined$Pr(Q)$ $\ =\$ undefined
$Pr(P \mid Q)$$\ =\$ $Pr(P\wedge Q)$ $/$ $Pr(Q)$ $\ =\$ undefined / undefined$\ =\$ undefined
$Pr(Q \mid P)$$\ =\$ $Pr(Q\wedge P)$ $/$ $Pr(P)$ $\ =\$ undefined / undefined$\ =\$ undefined
$Pr(P \mid P\vee Q)$ $\ =\$ $Pr(P\wedge (P\vee Q))$ $/$ $Pr(P\vee Q)$ $\ =\$ undefined / undefined$\ =\$ undefined
$Pr(Q \mid P\vee Q)$ $\ =\$ $Pr(Q\wedge (P\vee Q))$ $/$ $Pr(P\vee Q)$ $\ =\$ undefined / undefined$\ =\$ undefined