An interval scaled by a constant is an interval

[Rafael-SOWNet]

#322's arithmetic half.

Sumf and Minusf had interval cases and Mulf and Divf had none, so (0; 1) + 1 answered (1; 2) while (0; 1) * 2 was handed back. Three rows of Core/Sets/Arithmetics recorded that asymmetry as the expected behaviour, sitting directly beneath the two addition rows that answer:

[InlineData("[2; 3] + 3", "[5; 6]")]        // answers
[InlineData("3 * [2; 3]", "3 * [2; 3]")]    // expects itself back

The sign is the whole of it

factor	result
k > 0	ends and openness carried where they were
k < 0	reflected — ends swap, and their openness swaps with them
k = 0	left to the existing arm, see below
sign unknown	left alone

	Was	Is
3 * [2; 3]	3 * [2; 3]	[6; 9]
[2; 3] / 2	[2; 3] / 2	[1; 3/2]
[2; 3) * (-1)	unevaluated	(-3; -2]
(2; 3] / (-1)	unevaluated	[-3; -2)
0 - [0; 1)	-[0; 1)	(-1; 0]

The reflection is the half a test on the printed form alone would not catch: swapping the ends and leaving the flags where they were gives the right width and the wrong bracket at both ends. Membership is asserted too, for that reason.

That last row is not a subtraction at all — 0 - x is negated by an earlier arm, so it reaches Mulf as -1 * [0; 1). It is the case #1117 had to explicitly leave out, and it comes back for free.

An unknown sign is answered by not answering

(0; 1) * k for symbolic k is one interval when k is positive and the reflected one when it is negative. Picking either would be choosing which, so it is left unevaluated — which is what an unevaluated node means.

Two boundaries, asserted rather than left to be discovered

• (0; 1) * 0 still answers the number 0 rather than the set { 0 }. That arm is over every Entity, not only intervals, and moving it would change matrices and finite sets with it.
• 2 / (0; 1) is left alone. A constant over an interval straddling zero is two unbounded pieces rather than one interval, so there is no Interval to answer with.

On the issue's stated blocker

#322 says it "will be possible once we implement quantifiers". Scaling needs none — it is monotone in the factor's sign and in nothing else. Finite sets already map arbitrary functions (ln({1,2,3}) → {0, ln(2), ln(3)}), and this closes the interval-arithmetic gap.

What remains of #322 is applying a non-monotonic function to an interval: ln((0; 1)) is (-oo; 0) because ln increases, but sin((0; 7)) needs to know where the turning points are. That is a real piece of work and genuinely a proposal.

Recorded in BREAKING-CHANGES.md, both values measured on a build.

Full suite: 9059 passed, 14 skipped, 0 failed.

Part of #322. The issue is labelled Not now — this was implemented at @Rafael-SOWNet's direction after posting the measurement above; close it if the timing still stands.