I was going through the following slides and I wanted to show the following:

$$ \lambda x. x \equiv_{\alpha} \lambda y . y$$

formally. They define a an $\alpha$-conversion on page 15 as follows:

$$ \lambda x . E = \lambda z.(E[x \leftarrow z])$$

however, I wasn't sure how to formally show the statement I am trying to show. Essentially I guess I don't know how to formally show in a proof that two distinct objects actually belong to this same equivalence class. The intuition and idea is clear, but how do I know if I've shown the statement?

In fact if someone can show me how to do the more complicated one too that would be really helpful too:

$$ \lambda x.x (\lambda y . y) \equiv_{\alpha} \lambda y . y (\lambda x. x)$$

how do I know if I've shown what is being asked?

Actually I think page 16 is the one thats confusing me most:

Using the equation above, one has now the possibility to prove $\lambda$-expressions "equivalent". To capture this provability relation formally, we let $E \equiv_{\alpha} E^\prime$ denote the fact that the equation $E = E^\prime$ can proved using standard equational deduction form the equational axioms above (($\alpha$) plus those for substitution).

Exercise 3 Prove the following equivalences of $\lambda$-expressions:

  • $\lambda x.x \equiv_{\alpha} \lambda y.y$,
  • $\lambda x.x (\lambda y.y) \equiv_{\alpha} \lambda y.y (\lambda x.x)$,
  • $\lambda x.x(\lambda y.y) \equiv_\alpha \lambda y.y(\lambda y.y)$.

what does:

can be proved using standard equational deduction from the equational axioms above


Since there is already an answer that is not helping (because I don't understand the notation) I will add what I thought was the answer but I'm not sure:

I would have guessed that:

$$ \lambda x. x \equiv_{\alpha} \lambda y . y$$

if and only if there is a variable such that if we plug it into the lambda functions evaluates to the same function with the same variables. i.e.

$$ \lambda x. x \equiv_{\alpha} \lambda y . y \iff \exists z \in Var : \lambda x . x = \lambda z. ( (\lambda y . y)[y \leftarrow z] )$$

if we set $z = x$ we get:

$$\lambda z. ( (\lambda y . y)[y \leftarrow z] )$$ $$\lambda x. ( (\lambda y . y)[y \leftarrow x] )$$ $$\lambda x. (\lambda x .x )$$

which I assume the last line is the same as $\lambda x .x$ but I am not sure. If that were true then I'd show I can transform $\lambda y . y$ to $\lambda x . x$ which is what I assume the equivalence class should look like. Where did I go wrong?

  • $\begingroup$ The two expression $\lambda x.x$ and $\lambda x.(\lambda x.x)$ are definitely different. $\endgroup$ Oct 11, 2018 at 17:27
  • $\begingroup$ @YuvalFilmus what I would have assumed the right proof should look like is, ok I have one function and I can transform it to the other one by calling x = y, oh ok, look the functions now look exactly the same, so they must be in the same equivalence class. Thats what a correct proof I thought would look like... $\endgroup$ Oct 11, 2018 at 17:28

4 Answers 4


By definition of substitution we have $$x [x \leftarrow z] = z$$ therefore $$\lambda z . x [x \leftarrow z] = \lambda z . z \tag{1}$$ because $\lambda$-abstraction is a congruence (it preserves equality). By the definition of $\alpha$-equality we have $$\lambda x . x = \lambda z . x [x \leftarrow z] \tag{2}.$$ By transitivity of equality we get from (1) and (2) that $$\lambda x . x = \lambda z . z$$ If you require more details than this, you should use a computer proof assistant to check the details.

  • $\begingroup$ I think what I found confusing is the fact the definition of equivalence I would have used is, if there exists a variable substitution such that the lambda functions are syntactically equal, then they are equal. Instead they imposed the axiom $\lambda x . E = \lambda z. E[x\leftarrow z]$ as an axiom which seemed less natural to me. $\endgroup$ Oct 20, 2018 at 20:41
  • $\begingroup$ This is probably too much to ask but Id lve to see this through a theorem prover!!! :D $\endgroup$ Oct 20, 2018 at 20:45
  • $\begingroup$ Sorry, I don't really have time to formalize stuff like this. $\endgroup$ Oct 20, 2018 at 21:10
  • $\begingroup$ no need to apologize, thanks for the help Andrej! :) $\endgroup$ Oct 23, 2018 at 2:05

To be at least semi-formal, we can exploit these facts:

  1. $\lambda x.M \equiv_\alpha \lambda y.(M[x \leftarrow y])$ when $y$ is a variable not occurring free in $M$
  2. The relation $\equiv_\alpha$ is an equivalence relation. In particular, it is transitive, so we can perform the renaming of point 1. as many times as we want
  3. The relation $\equiv_\alpha$ is also a congruence, which means that "we can perform renaming in subterms as well". More formally, when $A \equiv_\alpha B$ then we have $M[x\leftarrow A] \equiv_\alpha M[x\leftarrow B]$ -- i.e., we can replace any occurrence of $A$ in a larger term with $B$, and the result will be $\alpha$ congruent.

So, for the exercises:

  • $\lambda x.x$ by point 1 can be rewritten as $\lambda y.(x[x\leftarrow y])$ which by definition of substitution is $\lambda y.y$
  • $\lambda x.x(\lambda y.y)$ by point 1 can be rewritten renaming $x$ into $y$. We can do this since $y$ is not free in $x(\lambda y.y)$. So, we get $\lambda y.y(\lambda y.y)$. Then we can apply the symmetric result we have proven before, i.e. $\lambda y.y \equiv_\alpha \lambda x.x$, and get (by congruence, point 3) the wanted $\lambda y.y(\lambda x.x)$. Since we performed two renamings, we implicitly relied on transitivity (point 2).

Finally, the sentence

can be proved using standard equational deduction from the equational axioms above

simply means: this can be proved by replacing some subterms with equivalent terms, possibly many times.

"Equational reasoning/deduction" refers to the process of substituting in a larger formula some subformula with an equivalent one, as if the two formulas could be assumed to be "equal", in some sense. This approach applies whenever we are working with a congruence relation.


The operation $[x \gets z]$ is defined inductively on the structure of the expression it is applied to. In particular, one of the rules is $x[x \gets z] = z$. This immediately implies $\lambda x.x \equiv_\alpha \lambda y.x[x\gets y] = \lambda y.y$.

  • $\begingroup$ I know your trying to help but this isn't helping. I wrote what I think a correct proof should look like. Can you take a look at that? $\endgroup$ Oct 11, 2018 at 17:25
  • $\begingroup$ Perhaps somebody else would come to the rescue. $\endgroup$ Oct 11, 2018 at 17:27
  • $\begingroup$ what does "can be proved using standard equational deduction from the equational axioms above" mean?. Thanks for the help. $\endgroup$ Oct 11, 2018 at 17:29
  • 1
    $\begingroup$ An example of an equational axiom is $x[x\gets y]=y$. An equational deduction also allows substituting equals for equals inside an expression. $\endgroup$ Oct 11, 2018 at 17:32
  • $\begingroup$ it seems that in your answer you said "This immediately implies " which makes me feel that you skipped the step I was looking to see to understand what proof means in this system. Do you mind making it explicit? $\endgroup$ Oct 11, 2018 at 17:34

My confusion was that the definition is not that there exists a variable name such that the two functions are equal (which I believe is a better definition, probably equivalent to the one provided). Instead they impose the following equational axiom:

$$ \lambda x.E = \lambda z.(E[x \leftarrow z]) $$

So we have:

$$ \lambda x.x = \lambda y.(E[x \rightarrow y]) $$

by equational axiom. Then:

$$ \lambda y.x[x \rightarrow y] = \lambda y.y$$

by realizing that $E = x$ and that substitution of $x$ for $y$ results in $y$ i.e.

$$ E[x \rightarrow y] = x[x \rightarrow y] = y$$

which results in:

$$ \lambda x.x = \lambda y.(E[x \rightarrow y]) = \lambda y.x[x \rightarrow y] = \lambda y.y $$

all in one line.


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