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I'm computing the interval C given the intervals A and B that makes the following true: $\forall x_c \in C$ there is a $x_a \in A$ such that $x_a + x_c \in B$.

Example: $A = [100,200]$, $B = [300,400]$, $C = [300-200,400-100] = [100,300]$.

I was wondering if this kind of calculation had a specific name (difference? distance?)

Ѕᴀᴀᴅ
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Yann
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According to Interval arithmetic - Wikipedia:

Subtraction: $[x_1, x_2] - [y_1, y_2] = [x_1 - y_2, x_2 - y_1].$

In your notation - assuming that the condition that you state is sufficient, as well as necessary, in order for $x_c$ to belong to $C$ - we have $x_c \in C$ if and only if there exists $x_a \in A$ such that $x_a + x_c \in B.$ This condition on $x_a, x_c$ holds if and only if there exists $x_b \in B$ such that $x_a + x_c = x_b.$ That is, there exists $x_b \in B$ such that $x_c = x_b - x_a.$

That is, using Wikipedia's notation: $x_c \in C$ if and only if $x_c \in B - A$; so $C = B - A.$

As the operation is called subtraction, presumably its result can be called the difference between $A$ and $B,$ but I haven't actually got a reference for this.

Calum Gilhooley
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    Oh! right if A = [min_a, max_a], then -A = [-max_a, -min_a] (to respect [lowest,highest] order), so it makes sense that if B + A is [min_b + min_a, max_b + max_a], B-A = B + (-A) = [min_b - max_a, max_b + min_a]. Thanks! – Yann Apr 28 '20 at 22:02