This is a best-effort (in the absence of runnable code) at making your inner lambda pointfree.
list2?.filter((something: Something) =>
R.all(
R.pipe(
R.repeat(1),
R.ap([prop('property1'), R.prop('property2')]),
R.zipWith(R.flip(R.applyTo), R.map(R.equals, [something.property1, otherObject])),
R.reduce(R.and, true),
R.not
),
list1
)
)
I really wouldn't bother trying to make the outer lambda pointfree (I just don't think it'll be readable)
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