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The interval dependency problem is the loss of relationships between repeated occurrences of the same input. For $X=[0,1]$, interval subtraction gives $X-X=[-1,1]$: it permits $u-v$ with independent choices $u,v\in X$. But the function $x\mapsto x-x$ has actual image of a function $\{0\}=[0,0]$. Giving both operands the name $X$ does not tell an interval arithmetic calculation to enforce $u=v$.

For the polynomial $f(x)=x(1-x)$ on $X=[0,1]$, the natural interval extension isThis allows the impossible combination $x=1$ in the first factor and $x=0$ in the second, producing the spurious upper bound $1$. In the actual function, both occurrences use the same $x$. Completing the square gives $x(1-x)=1/4-(x-1/2{)}^2$, which lies between $0$ and $1/4$ on $[0,1]$; every value between these bounds occurs, by continuity and the intermediate value theorem. The maximum $1/4$ occurs at $x=1/2$, while either endpoint gives the minimum $0$.

Interval subdivision covers $X$ by smaller closed intervals $X_j$, computes a valid interval enclosure $F(X_j)$ on each, and takes their union. Every allowed input belongs to a piece, so its output belongs to that union. If one closed interval is wanted, take the smallest one containing the union; gaps between separate output pieces can then be lost. Splitting narrows the freedom given to repeated inputs, while increasing the number of evaluations.

For $f(x)=x(1-x)$, split $[0,1]$ at $1/2$. The natural interval extensions areTheir union is $[0,1/2]$. Four equal pieces sharpen the upper bound again:The union is now $[0,3/8]=[0,0.375]$, still wider than the actual image of a function $[0,1/4]$. These are exact rational number bounds: the excess width comes from the interval dependency problem, rather than roundoff error.

A natural interval extension evaluates a chosen expression by replacing its arithmetic operations with interval arithmetic operations, its constants with enclosing closed intervals, and any additional functions with valid interval enclosures. Each intermediate result must remain defined on its inputs. The resulting $F(X)$ encloses the actual image of a function $f(X)$, but its width depends on the expression used. Rump's INTLAB tutorial describes this inclusion principle and warns that repeated inputs can produce severe overestimation.^[ref]