# First-order logic without equality

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.

• 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....
– user14972
Apr 16 '13 at 8:36
• I don't want to add a new binary relation, but to get rid of one. Apr 16 '13 at 8:43
• If you want to be able to talk about functional relations, you will need equality. Apr 18 '13 at 4:25
• @Hurkyl, equivalence is necessary but not sufficient for equality. Two objects can be isomorphic but distinct, and isomorphism is also an equivalence. Jul 26 '19 at 14:29

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 http://www.academia.edu/949632/How_to_live_without_identity_--_and_why

• And sorry for having fouled up the first time around! Apr 18 '13 at 15:29

For finite $\mathcal{A}$, identity is first order definable in $\mathcal{A}$ if and only if $\mathcal{A}$ is asymmetric (has no non-trivial automorphisms).

• What kind of thing is $\mathcal A$ supposed to be here? Nov 9 '17 at 8:49
• A finite $\Sigma$-structure for any vocab $\Sigma$ Nov 9 '17 at 13:25
• x @Samuel: Then your claim is definitely not true. For example, consider $$\mathcal A=(\{1,2,3\},\{(1,1),(2,2),(3,3)\})$$ in the language containing a single binary predicate. Nov 9 '17 at 16:06
• Oh thank you, I guess I should have clarified that the structure can't already have equality in it. Nov 9 '17 at 17:51
• How about $$\mathcal A=(\{1,2,3\},\{(1,2),(2,3),(3,1)\})$$ then? Now we can define equality as $$x = y \iff \exists z(p(x,z)\land p(y,z))$$ Nov 9 '17 at 17:54