TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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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.

For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list

Statement

Powers are the natural powers of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ and finite products those of The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, both taken in the commutative monoid $(Z, \cdot, 1)$ of $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ . Call $p : r \to Z$ an injective list of primes when every $p_{i}$ is prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and $p_{i} = p_{j}$ forces $i = j$ (Injection, surjection, bijection).

Let $n \in Z$ with $n \geq 1$ and let $p : r \to Z$ be an injective list of primes such that every prime divisor of $n$ equals $p_{i}$ for some $i < r$ . Then, with $v_{q}$ as in The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ :

$n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ;
$v_{q} (n) = 0$ for every prime $q$ that is not among $p_{0}, \dots, p_{r - 1}$ ;
the exponents are determined by $n$ : if $e : r \to N$ and $n = \prod_{i < r} p_{i}^{e_{i}}$ , then $e_{j} = v_{p_{j}} (n)$ for every $j < r$ .

Clause 3 needs only injectivity of the list, not the covering hypothesis.

Facts & Assumptions

Given: The commutative monoid $(Z, \cdot, 1)$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ , Semigroup and monoid); and the property $Q (r)$ : "for every $n \geq 1$ and every injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ".

[L1]

$\prod_{i < 0} g_{i} = e$ and $\prod_{i < σ (t)} g_{i} = (\prod_{i < t} g_{i}) g_{t}$ ; the value depends only on the entries named (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity).

[L2]

$g^{0} = e$ and $g^{σ (k)} = g^{k} g$ ; and $g^{m + n} = g^{m} g^{n}$ for natural exponents in a monoid (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L3]

For a prime $p$ and nonzero $a$ : $p^{k} \geq 1$ for every $k$ , so $p^{k} > 0$ and $p^{k} \neq 0$ ; $a = p^{v_{p} (a)} a^{'}$ with $a^{'} \neq 0$ and $p ∤ a^{'}$ ; $p^{k} ∣ a$ exactly for $k \leq v_{p} (a)$ ; $v_{p} (a) \geq 1$ exactly when $p ∣ a$ ; $v_{p} (1) = 0$ ; and $v_{p} (p) = 1$ (For a prime $p$ and a nonzero integer $a$ : $p^{v_{p} (a)} ∣ a$ and $p^{v_{p} (a) + 1} ∤ a$ ; $p^{k} ∣ a$ holds exactly for $k \leq v_{p} (a)$ ; $v_{p} (a) \geq 1$ exactly when $p ∣ a$ ; $v_{p} (1) = v_{p} (- 1) = 0$ ; and $v_{p} (p) = 1$ , The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ ).

[L4]

For a prime $p$ and nonzero $a, b$ : $a b \neq 0$ and $v_{p} (a b) = v_{p} (a) + v_{p} (b)$ ( $v_{p} (a b) = v_{p} (a) + v_{p} (b)$ for nonzero integers $a, b$ , and $v_{p} (a + b) \geq \min {v_{p} (a), v_{p} (b)}$ whenever $a$ , $b$ and $a + b$ are all nonzero).

[L5]

Every integer $> 1$ has a prime divisor (Every integer $n > 1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime); every prime $u$ satisfies $u > 1$ , and every positive divisor of a prime $w$ is $1$ or $w$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L6]

Divisibility is reflexive and transitive; $d ∣ u$ means $u = d c$ for some $c$ (Divisibility is reflexive and transitive on $Z$ , and is linear: if $d ∣ a$ and $d ∣ b$ then $d ∣ a x + b y$ for all integers $x, y$ ; also $d ∣ a$ implies $d ∣ a c$ , $- d ∣ a$ and $d ∣ - a$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L7]

Induction on $N$ (The principle of mathematical induction).

[L8]

On $N$ : $m \leq n$ means $m + c = n$ for some $c$ (Order on the natural numbers); $σ (k) = k + 1$ and addition is commutative (Addition of natural numbers, Addition is commutative, The natural numbers $N$ (von Neumann)); $t < σ (t)$ and $i < σ (t)$ exactly when $i \leq t$ (On $N$ the order is membership: $m < n ⟺ m \in n$ ); $m < n$ exactly when $σ (m) \leq n$ (Discreteness: $σ (n)$ is the immediate successor).

[L9]

$Z$ is a commutative ring; its order is total, antisymmetric and transitive, is compatible with addition, and positives are closed under multiplication; $ι : N \to Z$ is injective and order preserving with image the nonnegative integers, $ι (0) = 0$ , $ι (1) = 1$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals, The integers form a totally ordered ring, Order on the integers, The naturals embed in the integers).

Proof

technique · induction

1.1L3L5L9

Distinct primes do not divide one another: if $q$ and $w$ are primes with $q ∣ w$ , then $q$ is a positive divisor of $w$ (as $q > 1 > 0$ ) other than $1$ , so $q = w$ . Consequently $v_{q} (w) = 0$ whenever $q \neq w$ are primes.

1.2baseL1L5L9

$Q (0)$ holds: an empty list contains no prime, so $n$ has no prime divisor at all; if $n > 1$ it would have one, so $n = 1$ , and $\prod_{i < 0} p_{i}^{v_{p_{i}} (n)} = 1 = n$ .

1.3ih

Let $t \in N$ and assume $Q (t)$ .

1.4L3choose

Let $n \geq 1$ and let $p : σ (t) \to Z$ be an injective list of primes containing every prime divisor of $n$ . Write $π : = p_{t}$ , $v : = v_{π} (n)$ , and fix $n^{'}$ with $n = π^{v} n^{'}$ , $n^{'} \neq 0$ and $π ∤ n^{'}$ .

2.1step 1.1L2L3L4L7L8

For a prime $w$ and $k \in N$ : $v_{w} (w^{k}) = k$ , and $v_{q} (w^{k}) = 0$ for every prime $q \neq w$ . Both sets of $k$ for which these hold contain $0$ , since $w^{0} = 1$ and $v_{u} (1) = 0$ for every prime $u$ ; and if they hold at $k$ then, $w^{k}$ and $w$ being nonzero, additivity gives $v_{w} (w^{σ (k)}) = v_{w} (w^{k} w) = v_{w} (w^{k}) + v_{w} (w) = k + 1 = σ (k)$ and $v_{q} (w^{σ (k)}) = v_{q} (w^{k}) + v_{q} (w) = 0 + 0 = 0$ . Induction finishes both.

2.2step 1.4L3L8L9

$n^{'} \geq 1$ : $π^{v} \geq 1 > 0$ and $n > 0$ , so $n^{'} \leq 0$ would make $π^{v} n^{'} \leq 0 < n$ ; hence $n^{'} > 0$ , and a positive integer is $\geq 1$ because $n^{'} = ι (j)$ with $j \neq 0$ , so $1 = σ (0) \leq j$ .

3.1step 2.1L1L3L4L7L8L9

Clause 3. Let $p : r \to Z$ be an injective list of primes, $e : r \to N$ , and $M : = \prod_{i < r} p_{i}^{e_{i}}$ ; we claim $M \geq 1$ , that $v_{p_{j}} (M) = e_{j}$ for every $j < r$ , and that $v_{q} (M) = 0$ for every prime $q$ off the list. Let $T$ be the set of $r \in N$ for which this holds for all such $p$ and $e$ . Then $0 \in T$ : the empty product is $1$ , which is $\geq 1$ , has $v_{q} (1) = 0$ for every prime $q$ , and imposes no condition on indices. Suppose $r \in T$ and let $p : σ (r) \to Z$ be injective with primes and $e : σ (r) \to N$ . Writing $M^{'} : = \prod_{i < r} p_{i}^{e_{i}}$ we have $M = M^{'} \cdot p_{r}^{e_{r}}$ , with $M^{'} \geq 1 > 0$ and $p_{r}^{e_{r}} \geq 1 > 0$ , so $M \geq 1 > 0$ and both factors are nonzero; additivity then gives $v_{u} (M) = v_{u} (M^{'}) + v_{u} (p_{r}^{e_{r}})$ for every prime $u$ . Taking $u = p_{j}$ with $j < r$ gives $e_{j} + 0 = e_{j}$ , since $p_{j} \neq p_{r}$ by injectivity; taking $u = p_{r}$ gives $0 + e_{r} = e_{r}$ , since $p_{r}$ is off the list $p_{0}, \dots, p_{r - 1}$ ; and taking a prime $u$ off the whole list gives $0 + 0 = 0$ . So $σ (r) \in T$ , and $T = N$ by induction.

3.2step 1.4step 2.2L5L6L8

The restriction $p ↾ t$ is an injective list of primes containing every prime divisor of $n^{'}$ . Indeed if $q$ is prime with $q ∣ n^{'}$ then $q ∣ n$ , since $n = π^{v} n^{'}$ ; so $q = p_{i}$ for some $i < σ (t)$ ; and $q \neq π = p_{t}$ , because $π ∤ n^{'}$ ; hence $i \neq t$ , and $i < σ (t)$ gives $i \leq t$ , so $i < t$ .

3.3step 2.1step 1.4step 2.2L3L4L8

For $i < t$ we have $v_{p_{i}} (n) = v_{p_{i}} (π^{v}) + v_{p_{i}} (n^{'}) = 0 + v_{p_{i}} (n^{'}) = v_{p_{i}} (n^{'})$ , using additivity on the nonzero factors $π^{v}$ and $n^{'}$ , and step 2.1 with $p_{i} \neq π$ , which holds by injectivity since $i < t$ .

4.1step 1.3step 2.2step 3.2

By step 1.3 applied to $n^{'}$ and $p ↾ t$ : $n^{'} = \prod_{i < t} p_{i}^{v_{p_{i}} (n^{'})}$ .

5.1step 1.4step 4.1step 3.3L1L9

Therefore $\prod_{i < σ (t)} p_{i}^{v_{p_{i}} (n)} = (\prod_{i < t} p_{i}^{v_{p_{i}} (n)}) \cdot π^{v_{π} (n)} = (\prod_{i < t} p_{i}^{v_{p_{i}} (n^{'})}) \cdot π^{v} = n^{'} π^{v} = n$ , so $Q (σ (t))$ holds.

6.1step 3.1step 1.2step 5.1L3L7discharge-induction∎

With step 1.2 as base, induction gives $Q (r)$ for every $r \in N$ , which is clause 1. Clause 2 follows because a prime $q$ off the list does not divide $n$ — otherwise it would be a prime divisor of $n$ and hence on the list — so $v_{q} (n) = 0$ ; and clause 3 is step 3.1.

Remarks

This is The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ read by exponents rather than by lists, not a further theorem. A list of primes with repetitions becomes an injective list with multiplicities, and the multiplicity of $p_{i}$ is exactly $v_{p_{i}} (n)$ ; uniqueness up to a permutation becomes clause 3, that the exponent vector is determined.
There is deliberately no product over all primes. The familiar notation $\prod_{p} p^{v_{p} (n)}$ abbreviates a product with finitely many factors different from $1$ , and an indexed product with finite support is not available at this point in the library — The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity multiplies a finite list. The honest form quantifies over a finite injective list covering the prime divisors of $n$ and adds clause 2, which says every prime off the list contributes exponent $0$ . Nothing is lost: any two such lists agree on the exponents by clause 3.
Why injectivity is a hypothesis. Without it clause 3 fails outright: with $r = 2$ , $p_{0} = p_{1} = 2$ and $(e_{0}, e_{1}) = (1, 1)$ or $(2, 0)$ , both products are $4$ , so the exponent vector is not determined. Injectivity is what makes $v_{p_{j}}$ read off one multiplicity rather than a sum of several.

For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For n≥1n≥1 and any injective list p:r→Zp:r→Z of primes containing every prime divisor of nn, one has n=∏i<rpi vpi(n)n=∏i<r​pivpi​​(n)​; the exponents are determined by nn, and vq(n)=0vq​(n)=0 for every prime qq outside the list

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list