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Let $P(zss)$ mean concat (map f zss) = f (concat zss). We use structural induction on lazy lists, including the bottom value and allowing arbitrary, possibly partial, inner lists.

For [], the left side is [] and the right side is f [] = []. For the bottom value $\perp$, both map f and concat are strict functions in the outer spine, and the supplied strictness of $f$ makes the right side $f\perp =\perp$ too.

For the cons case, assume $P(zss)$ and write the new head as xs. The defining equations and the induction hypothesis giveThe third equality is exactly the supplied list concatenation preservation law. No inspection of xs was required, so the argument includes partial inner lists and undefined elements.

This proves the identity for every finite outer spine ending in either [] or $\perp$. In the usual pure Haskell semantics, these functions are Scott continuous maps. An infinite outer list is obtained as a limit by finite approximation of a lazy list; both composite functions preserve that limit. Equality on all its finite approximations therefore gives equality on the infinite outer list. Equivalently, $P$ is an admissible predicate on lazy lists. ConsequentlyThe continuity assumption is the usual semantic assumption for definable functions here; an extension from finite to infinite inputs cannot be justified by ordinary finite induction alone.

For every accumulator $a$, pattern matching makes foldl s a undefined the bottom value. Now use structural induction on a finite list prefix ending in the bottom value. The induction step is foldl s a (x:ps) = foldl s (s a x) ps; the induction hypothesis applies to every accumulator, including s a x. Consequently every such finite approximation has result $\perp$.

An infinite lazy list is the limit of its finite prefixes, each interpreted as a finite approximation of a lazy list. Definable Haskell operations are Scott continuous maps, so the result on the infinite lazy list is the supremum of the results on these approximations. Every result is $\perp$, hence foldl s a xs = undefined on every infinite input spine, irrespective of whether s would inspect the accumulator.

For a concrete separation, choose s = (&&), a = True, and xs = False : repeat True. The left fold is $\perp$, whereas the right fold returns False after observing the first element: False && r does not demand r. It is reaching the end of the input spine, rather than strictness of the combining operation, that prevents this left fold from returning.

Fix s and put $R(xs,a)=\mathrm{foldr}(ks)\mathrm{id}xsa$. The right fold equations giveThese are exactly the left fold equations. Structural induction, with the induction hypothesis quantified over all accumulators, proves equality on finite terminating lists.

To include the other input spines, first observe that both applied expressions give $\perp$ on an undefined spine. The same induction proves equality on every finite prefix ending in $\perp$: the step k s x f a demands the continuation f when its result is demanded. Finally use that the two applied operations are Scott continuous maps, taking the limit over prefixes interpreted as a finite approximation of a lazy list. This covers infinite spines, on which both applied expressions are $\perp$, and proves the identity for all lists under the stated Haskell semantics. The intermediate right fold builds an accumulator-transforming function; it does not provide a terminating left fold of an infinite list.