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In the context of $\lambda$-calculus, I was thinking about whether or not $$Mx =_{\beta\eta} Nx \implies M =_{\beta\eta} N$$ if $x\notin FV(M)\cup FV(N)$. I have been around this issue for quite some time now, but didn't reach anywhere useful.

Is it the case or not?

MJD
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Alistair -L.
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  • Have you tried structural induction on $M$? Is there a reason you eschew $\alpha$-conversion? – dfeuer Oct 13 '13 at 01:47
  • According to some exam question I've got, it is equivalent to $\lambda x.x =_{\beta \eta} \lambda xy.xy.$ Hope this helps. –  Dec 17 '13 at 07:36

1 Answers1

1

This is true regardless of whether $x \in FV(MN)$.

Proof. By hypothesis:

$$Mx =_{\beta\eta} Nx \tag{1}$$

Then also:

$$(\lambda x.Mx) =_{\beta\eta} (\lambda x.Nx) \tag{2}$$

The justification for $(2)$ is that the sequence of (possibly partially reversed) $\beta\eta$-contractions that proves $(1)$ will also prove $(2)$ when applied to the subterm $Mx$ of $(\lambda x.Mx)$.

As an aside, in the formal theory of $\lambda\beta\eta$ (and also $\lambda\beta$) this fact is accounted for by the 'weak extensionality' inference rule $\xi$: \begin{align} M &\;= M' \\ \hline (\lambda x.M) &\;= (\lambda x.M') \tag{$\xi$} \end{align}

Since

$$ \lambda\beta\eta \vdash M = N \iff M =_{\beta\eta} N$$

you could in principle just cite rule $\xi$ to deduce $(2)$ from $(1)$.

Furthermore, by the $\eta$-rule: $$(\lambda x.Mx) =_{\beta\eta} M \tag{3}$$ $$(\lambda x.Nx) =_{\beta\eta} N \tag{4}$$

So by $(2)$, $(3)$, $(4)$ and the symmetry and transitivity of $=_{\beta\eta}$ we have:

$$M =_{\beta\eta} (\lambda x.Mx) =_{\beta\eta} (\lambda x.Nx) =_{\beta\eta} N$$

which concludes the proof. $\square$

Note that the $\eta$-rule is generally added to a $\lambda$-theory to make it extensional (instead of intensional), so the truth of your statement was to be expected.

Roy
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