I was doing some review of propositional logic from Enderton's book. In one section(pg. 26 of the 2nd edition), he explains the idea that given wffs $\sigma_1, \sigma_2, \cdots, \sigma_k$ and $\tau$, one can use truth tables to check whether or not $\{ \sigma_1, \cdots, \sigma_k \} \models \tau$. Then he gives some examples of how one might in an educated way avoid checking all combinations of assignments of the propositional variables used in the sigmas and in $\tau$. This part looks o.k to me.

But then Enderton says "The stronger the antecedent(the expression on the left side), the weaker the conditional." He then gives the examples \begin{align} (P \wedge Q) &\models P \\ (P \rightarrow R) &\models ((P \wedge Q) \rightarrow R) \\ (((P \wedge Q) \rightarrow R) \rightarrow S) &\models ((P \rightarrow R) \rightarrow S) \end{align} What does he mean by this? Which antecedent is stronger, the first or the third? Maybe a question to start out with is: what do "strong" and "weak" even mean here? Thanks for any help/clarification!




2 Answers 2


"Strength" here means "provability strength". It's somewhat loose terminology, but the idea is the following. Suppose $A,B,C$ are all sentences, and $A \vDash B$ (and let's assume $B \not\vDash A$). In this case, we would say "$A$ is stronger than $B$", i.e. "$A$ can at least prove everything that $B$ can, and more."

Question 1: What do you think the relative strength of $\neg A$ and $\neg B$ would be? Hopefully, it's not too hard to see that, when you add negations, you flip the relative strengths; so now $\neg B \vDash \neg A$, i.e. "$\neg B$ is stronger than $\neg A$."

Question 2: What do you think the relative strength of $A \vee C$ and $B \vee C$ is, where $C$ is just some new sentence? Well, this is a bit tricky, since $C$ could be a tautology, but all else equal, if $A$ is stronger than $B$, then $A \vee C$ is at least as strong as $B \vee C$.

Combine these two together: If $A$ is stronger than $B$, then $\neg B$ is stronger than $\neg A$. But then $\neg B \vee C$ is stronger than $\neg A \vee C$, i.e. $B \rightarrow C$ is stronger than $A \rightarrow C$, given the definition of "$\rightarrow$". Slotting in the appropriate sentence for $A,B,C$ above, hopefully Enderton's remarks are a little more clear.

That's the formal answer to your question. Perhaps, on an intuitive level, consider the following: Let $A$ be "It's raining" and let $B$ be "It's raining really hard". In this case, $B$ is strictly stronger than $A$: $A$ follows from $B$ but not vice versa. But now consider the conditionals, "If it's raining, I will bring an umbrella" and "If it's raining really hard, I will bring an umbrella." Intuitively, the latter seems to be claiming something weaker. That is, the second conditional is consistent with it raining and me not bringing an umbrella (perhaps it's only lightly raining); but the first conditional is not (even if it's lightly raining, I'll bring an umbrella).

  • $\begingroup$ Do you have some reference that defines such terms? The hypothesis $ B\not \vDash A$ seems to be frequently avoided and, in some cases, is preferably to say that A is strictly stronger than B over A is stronger than B. $\endgroup$ Jun 8, 2022 at 17:44

I guess "$\sigma$ is stronger that $\tau$" or "$\tau$ is weaker than $\sigma$" means that $\sigma\models\tau$ but $\tau\not\models\sigma$.

First line: $(P\wedge Q)$ is stronger than $P$.

Second line: The conditional $((P\wedge Q)\rightarrow R)$ is weaker than the conditional $(P\rightarrow R)$ because the antecedent $(P\wedge Q)$ is stronger than $P$.

The third line follows from the second in much the same way as the second line follows from the first.


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