The fifth rule of inference is the formal version of the principle of mathematical induction, which we used once already informally in showing that the Fibonacci numbers form an arithmetic set. My preferred way of thinking about the principle of mathematical induction is as the statement that every non-empty set of natural numbers has a least element.
To see why this minimum principle implies the rule above consider the set of natural numbers which, when substituted for all free occurrences of \({ V }\) in \({ P }\), yield a false statement. If there are any then there’s a least one. It can’t be 0 because \({ Q }\) is true. If it’s not zero then it’s the successor of some natural number. Call that number \({ x }\). So substituting \({ x ' }\) for \({ V }\) in \({ P }\) gives a false statement. But \({ x ' }\) was the least number with this property so substituting \({ x }\) would give a true statement. Substituting \({ x ' }\) for \({ V }\) in \({ P }\) is the same as substituting \({ x }\) for \({ V }\) in \({ R }\) though and we have \({ [ ∀ V . ( P ⊃ R ) ] }\). Substituting \({ x }\) for \({ V }\) in this, which we are allowed to do by one of logical rules of inference for quantifiers, would give a contradiction, so our assumption that there is an integer which, when substituted into \({ P }\) for \({ V }\) makes the statement false is incorrect. In other words, substituting any value for \({ V }\) gives a true statement. But that’s the same as saying that \({ ( ∀ V . P ) }\) is true. So the minimum principle implies the principle of mathematical induction.
The reverse implication works as well. Suppose we have a set of natural numbers with no least element. Let \({ P }\) be the statement that no natural number less than the value represented by \({ V }\) belongs to the set. This is vacuously true when 0 is substituted for \({ V }\). Suppose it’s true for some other value. Then this value does not belong to the given set. If it did then it would be the least element of the set because the statement \({ P }\) tells us that no smaller number belongs to the set. Since there is no least element this can’t happen so the value \({ V }\) is not in the set. But then all numbers smaller than the value \({ V ' }\) are not in the set so from \({ P }\) we can deduce \({ P }\) with \({ V }\) replaced by \({ V ' }\), i.e. the statement we previously called \({ R }\). So we now have \({ Q }\) and \({ [ ∀ V . ( P ⊃ R ) ] }\) and therefore, by the principle of mathematical induction, \({ ( ∀ V . P ) }\). But \({ P }\) is the statement that no number less than \({ V }\) belongs to the set. This holds with \({ V }\) replaced by any numerical expression, including \({ x ' }\), where \({ x }\) is a variable. So no number less than \({ x ' };o\) belongs to the set and in particular \({ x }\) does not belong to the set. This holds for all natural numbers \({ x }\) so no natural number belongs to the set, which must therefore be empty. We’ve just seen that a set of natural numbers with no least element is necessarily empty. An equivalent way to say this is that every non-empty set of natural numbers has a least element, which is our minimum principle.
The proof above is an informal one. Indeed it can’t help but be informal. Our language for arithmetic has no notation for sets of natural numbers. We’ve seen how to express particular sets in this language but that’s not sufficient for the minimum principle, which is a statement about all sets of natural numbers. So there’s no way within Peano arithmetic to state the minimum principle, let alone prove its equivalence to the principle of mathematical induction. Once we have a language which includes sets, like the one we’ll introduce in the next chapter, we can give a formal statement of the minimum principle.