Proof

Suppose $N_1$ and $N_2$ are both normal forms for $M$. Then, $N_1 \stackrel{{R}}{\leftrightarrow} M \stackrel{{R}}{\leftrightarrow} N_2$, so $N_1 \stackrel{{R}}{\leftrightarrow} N_2$. There must then be a common $Z$ such that $N_1~R~Z$ and $N_2~R~Z$. Since neither $N_1$ nor $N_2$ can be reduced, it follows that $N1 \equiv N_2 \equiv Z$. $\Box$

Recall that our goal is to check whether normal forms in the lambda calculus are unique. One way to do this is to show that $\rightarrow ^*$ has the Church-Rosser property.

For any relation $\Rightarrow$, if $\Rightarrow$ has the Church-Rosser property, then so does $\Rightarrow^*$. The proof of this is easy to see pictorially.

$$\begin{array}{ccccccccccc} M & \Rightarrow & & \Rightarr... ...w & & \Rightarrow & & \cdots & & & & & \\ \end{array}$$

If $M \Rightarrow^* N_1$ (the horizontal axis) and $M \Rightarrow^* N_2$ (along the vertical axis), we can break up both reductions into a sequence of individual steps. From the fact that $\Rightarrow$ is Church-Rosser, we can complete the diamond for each single step in the sequence. In the end, we get a complete rectangle between $N_1$ and $N_2$ and the corner opposite $M$ is the common term that we can obtain from $N_1$ and $N_2$.

Unfortunately, $\rightarrow$ does not have the Church-Rosser property, so we cannot use this strategy to establish that $\rightarrow ^*$ is Church-Rosser! Consider the term $(\lambda x. xx)((\lambda x.x)(\lambda x.x))$. We have two reductions possible:

• $(\lambda x. xx)((\lambda x.x)(\lambda x.x)) \rightarrow ((\lambda x.x)(\lambda x.x))((\lambda x.x)(\lambda x.x))$ (Outermost)

• $(\lambda x. xx)((\lambda x.x)(\lambda x.x)) \rightarrow ((\lambda x.xx)(\lambda x.x)$ (Innermost)

If we take the second option, we can then in one step get

$$(\lambda x.xx)(\lambda x.x) \rightarrow ((\lambda x.x)(\lambda x.x))$$

We can reach this term from the first option as well, but it requires two steps!

The solution lies in defining an alternative notion of one step reduction from the basic beta reduction $\rightarrow _\beta$ such that

1. This new reduction is Church-Rosser.

2. Its reflexive, transitive closure is equal to $\rightarrow ^*$.

We define this new reduction $\twoheadrightarrow$ as follows.

$$\begin{array}{cccc} M \twoheadrightarrow M & \displaystyle... ...bda x.M)N \twoheadrightarrow M'\{x \leftarrow N'\}} \end{array}$$

The last inference rule builds in $\beta$ reduction. The crux of the relation $\twoheadrightarrow$ is that it combines nonoverlapping $\rightarrow$ reductions in one parallel step. For instance, if both $M$ and $N$ can be reduced to $M'$ and $N'$, respectively, then $M N$ can be reduced to $M'N'$ in one step.

We claim, without proof, that $\twoheadrightarrow$ is Church-Rosser and that $\twoheadrightarrow ^*$ is the same as $\rightarrow ^*$. Assuming the claim, it then follows that $\rightarrow ^*$ is Church-Rosser and so the normal form for a term in the lambda calculus is unique, if it exists.