Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 76 / 1 / Solution

Write ${\to }^{\ast }$ for the reflexive closure of the transitive closure of a relation $\to$, and ${\leftrightarrow }^{\ast }$ for its generated equivalence relation. The diamond property requires every pair $a\to b,a\to c$ to admit a one-step join $b\to d\leftarrow c$. The Church-Rosser property requires a possibly longer join whenever $b{\leftrightarrow }^{\ast }c$. Equivalently, every pair of finite reductions with the same source has a common finite reduct. Identity steps may be included in the diamond convention without affecting the counterexample below.

Let $A=(\lambda z.z)y$ and consider the lambda term $M=(\lambda x.xx)A$. Contracting its outer beta-redex gives $AA$, whereas contracting the inner beta-redex gives $(\lambda x.xx)y$. The only one-step reducts of $AA$ are $yA$ and $Ay$; the other branch has the sole one-step reduct $yy$. These terms are different even under alpha equivalence. Thus one-step beta reduction has no diamond property. Both branches do reach $yy$ by finite beta reduction.

For the positive result, use parallel beta reduction $\Rightarrow$, with rulesStructural induction, renaming binders to avoid capture, proves the substitution lemma: $P\Rightarrow P^{\prime }$ and $Q\Rightarrow Q^{\prime }$ imply $P[x:=Q]\Rightarrow P^{\prime }[x:=Q^{\prime }]$. Define the complete development of a lambda term byand $(PQ{)}^{\bullet }=P^{\bullet }{Q}^{\bullet }$ when the application is not initially a root beta-redex. Induction on a parallel derivation, using the substitution lemma in the contracting case, provesConsequently any two parallel reducts have the common successor $M^{\bullet }$. The ordinary one-step relation is included in the parallel relation, and every parallel step expands into finitely many ordinary steps; their finite closures therefore coincide. A finite grid of parallel diamonds joins any two finite parallel reductions. Finally a finite zigzag of forward and backward reductions can be joined inductively using that confluence. Hence beta reduction has the Church-Rosser property, although its individual steps do not have the diamond property.