Arithmetic, min/max and the three bitwise combining operators take two
operands or more now and fold left, which is what the examples were
already writing. The first pair still goes through `binary`, so the rule
about which side decides the type is unchanged for every call that was
already legal, and each operand after it is checked against that type.
min and max fold their own way: every step puts both sides in slots, the
accumulated pick included, so three operands are two nested lets and each
is still evaluated exactly once. Reusing the previous `if` as an operand
of the next would have copied everything inside it.
Three things stay at two operands, each for its own reason. A chain of
remainders is not something anyone writes on purpose; a chain of shifts
would pass two counts that are each legal for the width and still shift
the value away entirely. And a single operand is refused rather than
guessed: there is no unary minus in this language -- the prelude writes
every negation as (- 0 n) -- and no reciprocal, so both say so and name
the form to write instead.
Joseph Ferano
255367c6dc