CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The inverse $E$ satisfies $E (a + b) = E (a) E (b)$

Statement

For all $a, b \in R$ ,

$E (a + b) = E (a) E (b) .$

Facts & Assumptions

Given: $a, b \in R$ .

[F1]

$E = L^{- 1}$ , so $L (E (y)) = y$ , and $L$ is injective (The integral exponential $E : R \to (0, \infty)$ as the inverse of $L$ ).

[L1]

$L (x y) = L (x) + L (y)$ for positive $x, y$ (The integral logarithm satisfies $L (x y) = L (x) + L (y)$ for all positive $x$ and $y$ ).

Proof

technique · direct

1.1

Since $E (a), E (b) > 0$ , the product law gives $L (E (a) E (b)) = L (E (a)) + L (E (b)) = a + b .$

F1L1

2.1

Also $L (E (a + b)) = a + b$ . Injectivity of $L$ therefore gives $E (a + b) = E (a) E (b)$ .

F1step 1.1∎

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

OpenStax, Calculus Volume 1, Section 6.7 (standard reference, not scraped)

The inverse EE satisfies E(a+b)=E(a)E(b)E(a+b)=E(a)E(b)