= Solution
Use <function extensionality>: show $f(x)=g(x)$ for every value $x$ of the argument type. In <lazy evaluation> semantics, “every value” includes the <bottom value> and partial values. For <list> arguments, <structural induction on lazy lists> supplies the empty, bottom and cons cases, with continuity justifying passage to infinite <lists>.
The original printed equation has the same <right fold> on both sides, including the same operator and seed; this is visible in the scan.<Functional programming, data structures and algorithms, procedural programming, and digital hardware examination paper by the University of Oxford (2005)>[Printed p. 2, Question 1(c), displayed equation]{ref} Thus its literal answer is
$$
\boxed{\operatorname{foldr}\ (\oplus)\ e=\operatorname{foldr}\ (\oplus)\ e
\quad\text{whenever the expression is well typed}.}
$$
For every input $xs$, both sides denote the very same expression `foldr (oplus) e xs`; equality follows by reflexivity. This remains true if its value is $\bot$. Neither termination nor <associativity> nor an identity law is required. The scan does not establish a different intended equation, so no replacement by a <left fold> is assumed.
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