Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can we do without equality in first order logic? I looked at some cases in which equality is essential and found that it seems enough to have inequality implicit in the variables. Let $\phi(x,y)$ be a formula not containing equality.

Instead of writing $(\forall x)(\exists y) x \neq y \wedge \phi(x,y)$ one can write $(\forall x)(\exists y) \phi(x,y)$, supposing that different variables denote different objects. Correspondingly:

$(\exists x)(\exists y) \phi(x,y)$ instead of $(\exists x)(\exists y) x \neq y \wedge \phi(x,y)$

$(\forall x)(\forall y) \phi(x,y)$ instead of $(\forall x)(\forall y) x = y \vee \phi(x,y)$

$(\exists x)(\forall y) \phi(x,y)$ instead of $(\exists x)(\forall y) x = y \vee \phi(x,y)$

The other way around, if we want to express that $(\forall x)(\exists y) \phi(x,y)$ allowing the case $x=y$, we have to write $(\forall x)(\exists y) \phi(x,x) \vee \phi(x,y)$, and correspondingly

$(\exists x)(\exists y) \phi(x,x) \vee \phi(x,y)$ instead of $(\exists x)(\exists y) \phi(x,y)$

$(\forall x)(\forall y) \phi(x,x) \wedge \phi(x,y)$ instead of $(\forall x)(\forall y) \phi(x,y)$

$(\exists x)(\forall y) \phi(x,x) \wedge \phi(x,y)$ instead of $(\exists x)(\forall y) \phi(x,y)$

Another case is "there is exactly one object $x$ with $\phi(x)$":

$(\exists x)(\forall y) \phi(x) \wedge \big(\phi(y) \rightarrow y = x\big)$ becomes $(\exists x) \phi(x) \wedge \neg (\exists x)(\exists y)\phi(x) \wedge \phi(y)$

I wonder if every sentence of first order logic including equality can be equivalently written without equality when interpreting different variables to denote different objects. For the time being I'd like to restrict the question to languages without function symbols and individual constants, i.e. comprising only relation symbols and variables.

share|cite|improve this question
There is no reason for different variables to denote different objects. Different variables are different objects but they can be interpreted as the same object if they are free in a formula. It depends of the model you choose. – leo Apr 16 '13 at 7:59
If you allow relation symbols, a theory can always just add a new binary relation symbol EQUALS and the axioms ensuring it is an equivalence relation.... – Hurkyl Apr 16 '13 at 8:36
I don't want to add a new binary relation, but to get rid of one. – Hans Stricker Apr 16 '13 at 8:43
If you want to be able to talk about functional relations, you will need equality. – Dan Christensen Apr 18 '13 at 4:25
That's why I restricted my question to languages without function symbols (to be on the safe side). On the other hand I tried to show how to circumscribe "there is exactly one x..." – Hans Stricker Apr 18 '13 at 9:12
up vote 15 down vote accepted

The question is this:

Can we do without equality in first order logic, and get something equally expressive using a language with the semantics for quantifiers tweaked so that different variables get assigned different values?

The proposal here in fact goes back to Wittgenstein's Tractatus 5.53, where he writes, ‘Identity of the object I express by identity of the sign and not by means of a sign of identity. Difference of the objects by difference of the signs.’ Can this proposal, not really developed out by Wittgenstein, be made to work?

The answer is it that it can, as shown by Hintikka in 1956 ('Identity, Variables, and Impredicative Definitions', Journal of Symbolic Logic). Hintikka distinguishes the usual 'inclusive' reading of the variables (i.e. we are allowed to assign the same object to distinct variables) from the 'exclusive' reading, and then proves the key theorem (summarized on p. 235):

[E]verything expressible in terms of the inclusive quantifiers and identity may also be expressed by means of the weakly exclusive quantifiers without using a special symbol for identity.

So yes, Hans Striker's conjecture is right. For a more recent revisiting of Hintikka's result, in the context of interpreting the Tractatus see e.g. Kai F. Wehmeier's 'How to Live without Identity - And Why', Australasian Journal of Philosophy 2012, downloadable at

share|cite|improve this answer
Peter: Thank you very much! I am very happy with this answer. – Hans Stricker Apr 18 '13 at 15:17
And sorry for having fouled up the first time around! – Peter Smith Apr 18 '13 at 15:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.