I am working on $\lambda$-terms and trying to prove the $=$ is an equivalence relation on $\lambda$-terms. My problem is proving reflexive relation.

$\frac{}{\theta \vdash x = x}$

$\frac{ \theta,x \# N, y \# M \vdash M = [y := x]N } { \theta \vdash \texttt{$\lambda x.M$ $=$ $\lambda y.N$} } $

$\frac{\theta \vdash M_1 = M_2 \quad N_1 = N_2 } { \theta \vdash \texttt{$M_1 \, N_1$ $=$ $M_2 \,N_2$} }$

I put two restrictions on such $\lambda$-terms.First, bound variables are distinct. For example, there are no two terms such as $\lambda x.M$ and $\lambda x.N$. Second, multiple bindings of a variable is not allowed. For example, $\lambda x. \lambda x.M$ should be written as $\lambda y. \lambda x.M$. To sum up, every bound variable should be distinct. $[y := x]N$ means variable $x$ replaces variable $y$ in term $N$. $x \# N$ means $x$ does not occur in $N$. $\theta$ is a set of $\#$. Also, $\alpha$-equivalence is assumed for the terms. For example, $\lambda x.x = \lambda y.y$

I tried to prove that the $=$ shown in above rules is an equivalence relation on such terms.

For equivalence relation, I have to prove the following three relations.

reflexive: $\theta \vdash M=M$.

symmetric : $\theta \vdash M=N$ implies $\theta \vdash N=M$.

transitive: $\theta \vdash M=N$ and $\theta \vdash N=P$ implies $\theta \vdash M=P$.

The proof of reflexive relation is the following.

when $M$ is a variable such as $x$, then $x = x$.

when $M$ is an application such as $M_1 \,N_1$), then I have $M_1 \,N_1$ = $M_1 \,N_1$, so it is true.

when $M$ is an abstraction such as $\lambda x.M$, from $\lambda x.M = \lambda x.M$, I have $ x \# M \vdash | M=[x:=x]M $, which is not true becuase $x \in M$. Also, as I said, there are no two terms such as $\lambda x.M$ and $\lambda x.M$, so I cannot show $M=M$ for an abstraction.

Since $\alpha$-equivalence is assumed for terms. I assume that $M=\lambda x.M_1 =\lambda y.M_2$. Therefore, I will have $x \# M_2, y \# M_2 \vdash M_1=[y:=x]M_2$? is this the right way to prove reflexivity?

I would appreciate your kind help.

  • $\begingroup$ It seems impossible to me to prove $\lambda x.x = \lambda x.x$ with those restrictions. Can you actually prove that? $\endgroup$
    – chi
    Commented Mar 15, 2017 at 12:36
  • $\begingroup$ @chi I think I can't write $\lambda x.x = \lambda y.y$ because of the restriction. Any idea? $\endgroup$
    – alim
    Commented Mar 15, 2017 at 13:32
  • 1
    $\begingroup$ I'm quite unsure about this. It seems that reflexivity fails on $\lambda x.x$. If so, the system can not be reflexive, unless the restrictions are relaxed or the definition is adapted somehow. $\endgroup$
    – chi
    Commented Mar 15, 2017 at 13:49
  • $\begingroup$ @chi how about write $\lambda y.y =  \lambda x.x$ since both are the same lambda term. Since $\lambda x.x$ represents a class of terms. Do you think this is possible? $\endgroup$
    – alim
    Commented Mar 15, 2017 at 14:01
  • 2
    $\begingroup$ Yes that should be provable, one starts from $x=x$ which is the same as $x=[y:=x]y$. Then, since $x\# y$ and $y\# x$ one can have $\lambda x.x = \lambda y.y$. If you consider $\alpha$-convertible terms as equal, then the relations above might be reflexive, after all (I'm unsure at the moment). $\endgroup$
    – chi
    Commented Mar 15, 2017 at 14:33

1 Answer 1


I will only prove reflexivity $A=A$. We proceed by induction on the structure of $A$.

If $A\equiv x$, then $x=x$ follows from rule 1.

If $A\equiv \lambda x. M$, then by induction hypothesis we can assume $M=M$ is provable. Now, take any $y$ not free in $M$. We have $M \equiv [y:=x][x:=y]M$ by construction. Let $N \equiv [x:=y]M$. Rule two states

$$ \dfrac{ x\# N, y\# M \vdash M=[y:=x]N }{ \vdash \lambda x.M = \lambda y.N } $$

We indeed have $x\# N$, since the $x$ variable was renamed as $y$ in $N \equiv [x:=y]N$. We also have $y\# M$ by our choice of $y$. Induction hypothesis states $M=M$, which is $M=[y:=x]N$. Hence, the conclusion of the rule holds, which is $\lambda x. M = \lambda y. [x:=y]M$. If terms are identified up-to $\alpha$, then this equality is the same as $\lambda x.M = \lambda x.M$.

If $A=MN$, the induction hypotheses and rule 3 suffice.

  • $\begingroup$ I usually follow my intuition, and I am not good at presenting them. With discussing with you above and after reading your proof, I clarified some points which were not clear to me. Thank you, helped me a lot. $\endgroup$
    – alim
    Commented Mar 18, 2017 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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