TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-01

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 exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$

Statement

For all real $x, y$ , $\exp (x + y) = \exp (x) \exp (y) .$

Facts & Assumptions

Given: $x, y \in R$ .

[L1]

For fixed $x, y$ , the auxiliary power series $\sum_{n \geq 0} x^{n} z^{n} / ι (n!)$ and $\sum_{n \geq 0} y^{n} z^{n} / ι (n!)$ have infinite radius by The exponential series converges absolutely for every real argument. Inside their common radius, their product is the Cauchy product of their coefficients (Inside the common radius the product of two power-series sums is represented by the Cauchy product of their coefficients).

[L2]

The binomial theorem says $(x + y)^{n} = \sum_{k = 0}^{n} ι (\binom{n}{k}) x^{k} y^{n - k}$ (The binomial theorem in $R$ : $(x + y)^{n} = \sum_{k < n + 1} ι ⁣ (\binom{n}{k}) x^{k} y^{n - k}$ , The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L3]

For $k \leq n$ , $ι (\binom{n}{k}) = ι (n!) / (ι (k!) ι ((n - k)!))$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ). Therefore $1 / (ι (k!) ι ((n - k)!)) = ι (\binom{n}{k}) / ι (n!)$ , with all naturals read in $R$ through The canonical natural $ι (n) = n \cdot 1_{F}$ of a field.

Proof

technique · direct

1.1L1given

Apply [L1] at the auxiliary value $z = 1$ . The coefficient of degree $n$ in the resulting Cauchy product for $\exp (x) \exp (y)$ is $\sum_{k = 0}^{n} x^{k} y^{n - k} / (ι (k!) ι ((n - k)!))$ .

2.1step 1.1L2L3algebra

Apply [L3] and [L2] to identify this finite sum with $(x + y)^{n} / ι (n!)$ .

3.1step 2.1L1∎

Summing over $n$ gives the exponential series at $x + y$ , hence the formula.

The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y)exp(x+y)=exp(x)exp(y)

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$