I am reading a logic book in my free time and usually the inference rule of universal generalization is motivated by real-life examples: Imagine having the statement that all people with brown hair are tired in the morning and that all people who are tired in the morning like coffee. Then one can prove that everyone with brown hair likes coffee: Let x denote a person that has brown hair, then x is tired in the morning and thus x likes coffee. Since we did not assume anything about the object (in this case person) x, it seems reasonable that for every object (person) x the reasoning should hold. If it did not, that would mean that there would have to exist an object y that does not satisfy this property, but then we could just do the same proof for this very y.

Now, when moving on to abstract objects (such as natural numbers for simplicity), this is more difficult to me. I suspect this is due to the abstraction of mathematical objects (what should they be?). When proving something about all natural numbers, it is again common to let n denote any object that satisfies being an element of the natural numbers and then deduce the statement. Since we assume the inference rule of universal generalization, this then holds for all natural numbers.

Question: However, why is it reasonable to assume that universal generalization holds for natural numbers (or more generally abstract objects)? Are there any good viewpoints on this? I assume that the "obvious" one is to just follow the real life example, but I am curious if there are any other views on this, as I seem to struggle with finding better ones quite a bit.

  • To avoid this difficulty, think of it in terms of manipulated symbols, not what they represent. Universal generalization works because arguing with not otherwise specified symbol simply unfolds the rules for manipulating formulas with ∀ into a more parsed form. What the symbols happen to represent, concrete, abstract or nothing at all, makes no difference whatsoever. For justifying the analogy of numbers to concrete objects and applying ∀ to them see abstract objects, but that has little to do with universal generalization in particular.
    – Conifold
    Commented Dec 29, 2021 at 22:31
  • @Conifold Thank you for your comment, I would appreciate if you would expand it to an answer, since I think I would benefit from a more detailed version of your comment. Especially, if you could mention what you mean by manipulated symbols, explain why it makes no difference whatsoever and giving more insight into your last sentence, I would be very greatful. Commented Dec 29, 2021 at 23:14
  • Because in final terms, all objects are abstract; objects only exist in reason, not necessarily physically. Descartes' cogito ergo sum means not that he's a physical object (in fact, he doubts about the existence of an external world and its things), but that he perceives himself as an abstract object, by means of thinking, therefore he exists... for himself.
    – RodolfoAP
    Commented Feb 2, 2022 at 4:18
  • 1
    Your suspect of UG's applicability in non-trivial abstract theories such as the famous Robinson arithmetic Q is not wrong. Surprisingly due to nonstandard models per Godel's incompleteness Q⊬∀x(0+x=x), though Robinson Q can easily prove ∀x∈N(0+x=x)! Indeed for such inductively axiomatized abstract system, for UG to work we need to add an infinitary ω-rule of ω-logic, then per model theory's omitting type theorem a theory has an ω-model iff it is consistent in ω-logic... Commented Feb 2, 2022 at 5:19
  • 1
    @NoahSchweber thx for your critique for my above comment made 5 months ago! In Q ∀x(x+0=x) is axiom 4, but a famous result is Q⊬∀x(0+x=x) by simply adding 2 nonstandard members satisfying certain relations which can quickly demo incompleteness of Q and surprising for people not familiar with 1st order arithmetic incompleteness (generic group theory incompleteness is easier to understand for Abelian/non-Abelian models). Thus for natural numbers satisfying commutativity I said it's easy to prove ∀x∈N(0+x=x). The point is in Q/PA it's more than natural numbers thus UG needs applied with care... Commented Jul 3, 2022 at 4:55

2 Answers 2


Here is another way of looking at the matter. Let a,b,c... be of type "people with brown hair". Let x,y,z.... be of type "natural number". Lets use your coffee example:

Say that you have some algorithmic procedure by which you can produce a proof that any x, that is, any natural, has some property P; further such a procedure does not rely on any specific property of x, except for the fact that x is of type natural. This is akin to your stipulation that any person with brown hair likes coffee- it doesn't rely on any fact about a specific person, only on their "brown hair -ness ".

Since the purported proof relies only on the properties of being natural, and not any specific property of any specific natural, it may be duplicated for any natural number. Hence, you essentially have a proof that every natural has P.

The situation is entirely analagous to the coffee example given above, except here, having brown hair means having the property of being a natural number.


By "universal generalization", I assume you mean the rule that if you can derive P(x), then you can infer forall x.P(x). There are two ways to understand this. First, you can view it as a formal property of the logic. If a logic allows you to introduce and reason with free variables, then it needs a rule to say what a proposition with a free variable means. The rule needn't have any meaning beyond that. For example, some calculuses have a rule of renaming that basically says you can change the name of a variable so long as you do it in a certain way. This is purely formal, and has no higher meaning.

What universal generalization does is cause all unbound variables to be implicitly universally quantified. Without universal generalization it wouldn't be clear what P(x) means when x is an unbound variable (that is, when x is not a constant and is not bound by a quantifier. In this sense, the rule is just a formal property of the logic.

An alternative way to view this topic is logically. In this case it is possible to prove that universal generalization is valid. Suppose you have a valid proof P in which x occurs as a free variable, concluding with well-formed formula C(x) in which x occurs as a free variable, then it must be the case that forall x.C(x).

Proof (by contradiction): suppose that there is some object a such that not C(a). Then replace x in the proof with a, and you have a new proof proving that C(a), which is a contradiction.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .