@PeterTaylor’s excellent answer points to exactly the result wanted. But the proof there can be streamlined a bit, and may be paywalled for some readers, so I’ll write it out here for accessibility.

**Theorem.** Let $\newcommand{\B}{\mathbf{B}}\B$ be any set of combinators of rank ≤2. Then there is no $\B$-expression $T$ such that $\newcommand{\reddto}{\twoheadrightarrow}Tabc \reddto cab$.

**Proof.** Fix such $\B$ throughout. The idea is (unsurprisingly) to find an invariant of $Tabc$ that is sufficiently closed under reduction, and not satisfied by $cab$. Say that an expression is *bad* if it is of the form $E[F[a,b],c]$, where $E$ and $F$ are each $\B$-expressions of 2 variables. Then we have:

**Lemma.** Any bad expression is either normal, or else has a reduction chain to another bad expression, including at least one leftmost reduction.

**Proof of lemma.** If a bad expression $E[F[a,b],c]$ is not stuck, consider its leftmost redex. There are three possible cases:

The head combinator of the redex is not in an occurrence of $F$. Then each occurrence of $F$ is either outside the redex, or inside an argument of the head combinator; so the reduct is of the form $E'[F[a,b],c]$.

The whole redex is inside an occurrence of $F$. Then by reducing it in *every* occurrence of $F$, we have $E[F[a,b],c] \reddto E[F'[a,b],c]$, and this includes the leftmost redex.

The head combinator $H$ is iside an occurrence of $F$, but not the whole redex. Then $H$ must be of rank ≥1. If $H$ is of rank 1, then $F$ is just $H$, so $x,y$ don’t occur, and the whole expression (both before and after reduction) is of the form $E'[c]$, so is still bad. If the head combinator is of rank 2, then we must have $F[a,b] \equiv HF'[a,b]$, so by re-parsing as $E[F[a,b],c] \equiv E'[F'[a,b],c]$, case 1 applies.

This proves the lemma. It follows that any bad expression either reduces to a bad normal form, or has an infinite quasi-leftmost reduction sequence. So by the quasi-leftmost-reduction theorem (Hindley–Seldin 2008, Thm 3.22), any normal form of a bad expression is bad; and so the bad expression $Tabc$ cannot reduce to the normal, non-bad expression $cab$. □

References:

- Rémi Legrand,
*A Basis Result in Combinatory Logic*, J. Symb. Logic 1988; jstor full text
- J. Roger Hindley, Jonathan P. Seldin,
*Lambda-Calculus and Combinators, an Introduction*, CUP 2008

combinator of rank $n$is one with a reduction rule $C x_1 \ldots x_n \to E$ where $E$ is formed just from $x_1,\ldots,x_n$ and application. So more complex combinators like the universal $\iota$ don’t fit that form. It would be nice to have a more general definition of rank that extended Legrand’s result to cover $\iota$ and similar, but I’ve not come across such a definition. $\endgroup$3more comments