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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.

A lambda term is built recursively as a variable $x$, an application $MN$, or a lambda abstraction $\lambda x.M$. Application associates to the left and abstraction binds occurrences of its variable. We identify consistent renamings of bound variables by alpha equivalence. A beta reduction contracts $(\lambda x.M)N$ to the capture-avoiding substitution $M[x:=N]$, at any subterm. Write ${\to }_{\beta }^{\ast }$ for finitely many such steps, including none. Beta equivalence is the reflexive, symmetric, transitive closure of reduction, so it is represented by a finite zigzag of forward and backward steps.

To prove the Church-Rosser theorem, introduce parallel beta reduction $\Rightarrow$ by the four rulesBy structural induction every term parallel-reduces to itself. Each ordinary step is a parallel step. Conversely, each parallel step can be performed by an ordinary finite sequence: reduce the displayed subterms recursively, and in the fourth rule contract the root last. Hence ${\Rightarrow }^{\ast }$ and ${\to }_{\beta }^{\ast }$ are the same relation.

The substitution lemma isProve it by induction on the parallel-reduction derivation for $M$. The variable case separates $M=x$ from other variables; application and abstraction follow from their inductive hypotheses. For abstraction, rename its binder fresh for $x,N,N^{\prime }$ before substituting. For a contracted redex, choose its binder $y$ distinct from $x$ and fresh for $N,N^{\prime }$. The substitution identity $P[y:=Q][x:=N]=P[x:=N][y:=Q[x:=N]]$ follows by structural induction with fresh binders; the variable cases use $y\notin FV(N)$, and application and abstraction preserve the identity. Apply the hypotheses to the body and argument, then the parallel contraction rule and this identity. This proves the lemma for every derivation, including simultaneous contractions inside a redex.

Define the complete development of a lambda term byOnly a root redex already present in the source application is contracted by this clause; a newly created root redex need not be contracted.

We claim that $M\Rightarrow N$ implies $N\Rightarrow M^{\bullet }$. Induct on the derivation. The variable and abstraction cases are immediate. In the ordinary application rule, use the inductive developments of both components. If its original function is an abstraction, also contract the resulting root redex by the parallel contraction rule; otherwise just use the application rule, even if the reduct has acquired a new root redex. In the contraction rule, the reduct has form $P^{\prime }[x:=Q^{\prime }]$, and the substitution lemma with $P^{\prime }\Rightarrow P^{\bullet }$, $Q^{\prime }\Rightarrow Q^{\bullet }$ gives $P^{\prime }[x:=Q^{\prime }]\Rightarrow P^{\bullet }[x:=Q^{\bullet }]$. This establishes the claim.

Consequently, any two parallel reducts $N,P$ of $M$ each parallel-reduce to $M^{\bullet }$. Thus parallel reduction has the one-step diamond property. Its finite transitive closure is confluent: fill a finite grid of elementary diamonds to join two finite reduction paths from the same term. Since its closure equals ordinary finite beta reduction,A finite conversion zigzag now also has a common reduct, by induction along it. If the next zigzag step points forward from a term already having a common reduct, join its two outgoing paths by confluence; if it points backward, compose the new forward path to that common reduct. The reverse implication is immediate because two paths to a common reduct form a conversion zigzag. This proves

A beta-normal form contains no beta-redex, so its only finite reduct is itself. If two normal forms are beta-equivalent, their common reduct must equal both. Hence a term has at most one beta-normal form up to alpha equivalence. The qualification about bound-variable names is essential.

For an example without normal form, take the Omega combinatorIts only redex is the root, and contracting it reproduces $\Omega$. Every finite reduct is therefore $\Omega$, which is not a beta-normal form. By the theorem it cannot even be beta-equivalent to any normal form.