LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$

Statement

Let $a, b : N \to N$ , let $c \in N$ , and let $m, n \in N$ , with $\sum^{N}$ and $\prod^{N}$ as in Finite sums and finite products of natural numbers, $\sum_{k < n} a_{k}$ and $\prod_{k < n} a_{k}$ in $N$ and $\sum^{R}$ , $\prod^{R}$ as in Finite sums and finite products, by recursion. Let $ι : N \to R$ be the canonical natural of The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, so $ι (0) = 0$ and $ι (σ (n)) = ι (n) + 1$ . Then:

$ι$ is additive and multiplicative. $ι (1) = 1$ , and $ι (m + n) = ι (m) + ι (n)$ and $ι (m n) = ι (m) ι (n)$ for all $m, n \in N$ , the cases where a factor is $0$ included.
Additivity. $\sum_{k < n} (a_{k} + b_{k}) = \sum_{k < n} a_{k} + \sum_{k < n} b_{k}$ .
Constants. $\sum_{k < n} c = n \cdot c$ , the summand being the constant list.
Splitting. If $m \leq n$ and $d : = n - m$ , then $\sum_{k < n} a_{k} = \sum_{k < m} a_{k} + \sum_{j < d} a_{m + j}$ , and $\prod_{k < n} a_{k} = (\prod_{k < m} a_{k}) (\prod_{j < d} a_{m + j})$ .
Monotonicity. If $a_{k} \leq b_{k}$ for every $k < n$ then $\sum_{k < n} a_{k} \leq \sum_{k < n} b_{k}$ ; and $a_{j} \leq \sum_{k < n} a_{k}$ for every $j < n$ .
Products. $\prod_{k < n} (a_{k} b_{k}) = (\prod_{k < n} a_{k}) (\prod_{k < n} b_{k})$ ; and if $a_{k} \neq 0$ for every $k < n$ then $\prod_{k < n} a_{k} \neq 0$ .
The bridge into $R$ . $ι (\sum_{k < n}^{N} a_{k}) = \sum_{k < n}^{R} ι (a_{k})$ and $ι (\prod_{k < n}^{N} a_{k}) = \prod_{k < n}^{R} ι (a_{k})$ .
$ι$ is strictly increasing, hence injective. $m < n$ if and only if $ι (m) < ι (n)$ , and $m = n$ if and only if $ι (m) = ι (n)$ .

Clauses 6 and 7 together are the licence used everywhere below: an identity between natural numbers may be proved by proving the corresponding identity between their canonical naturals in $R$ , and conversely a real identity whose two sides are canonical naturals is an identity in $N$ .

Facts & Assumptions

Given: Lists $a, b : N \to N$ , a natural $c$ , naturals $m, n$ , and the ambient ordered field $R$ . Recall $σ (n) = n + 1$ and the truncated difference $n - m$ of Finite sums and finite products of natural numbers, $\sum_{k < n} a_{k}$ and $\prod_{k < n} a_{k}$ in $N$ .

[L1]

Induction: a property holding at $0$ and inherited by successors holds at every natural (The principle of mathematical induction).

[L2]

Recursion clauses in $N$ (Finite sums and finite products of natural numbers, $\sum_{k < n} a_{k}$ and $\prod_{k < n} a_{k}$ in $N$ ): $\sum_{k < 0} a_{k} = 0$ , $\sum_{k < σ (n)} a_{k} = \sum_{k < n} a_{k} + a_{n}$ , $\prod_{k < 0} a_{k} = 1$ , $\prod_{k < σ (n)} a_{k} = (\prod_{k < n} a_{k}) \cdot a_{n}$ .

[L3]

Recursion clauses in $R$ (Finite sums and finite products, by recursion): $\sum_{k < 0}^{R} x_{k} = 0$ , $\sum_{k < σ (n)}^{R} x_{k} = \sum_{k < n}^{R} x_{k} + x_{n}$ , and likewise $\prod_{k < 0}^{R} x_{k} = 1$ , $\prod_{k < σ (n)}^{R} x_{k} = (\prod_{k < n}^{R} x_{k}) \cdot x_{n}$ .

[L4]

Arithmetic of $N$ : addition and multiplication are associative and commutative, $0 + n = n$ and $n + 0 = n$ , $1 \cdot n = n \cdot 1 = n$ and $0 \cdot n = n \cdot 0 = 0$ , multiplication distributes over addition, and $σ (a) \cdot n = a \cdot n + n$ (Addition is associative, Addition is commutative, Left identity for addition, Multiplication is associative, Multiplication is commutative, Zero and one under multiplication, Distributivity and the successor law for multiplication, Addition of natural numbers, Multiplication of natural numbers).

[L5]

Order of $N$ : $m \leq n$ means $m + j = n$ for some $j$ , that $j$ is unique, $m \leq n ⟺ m + k \leq n + k$ , and $m < n ⟺ σ (m) \leq n$ , so $n \neq 0$ is the same as $1 \leq n$ ; exactly one of $m < n$ , $m = n$ , $n < m$ holds (Order on the natural numbers, Addition is cancellative, Order is compatible with addition, Discreteness: $σ (n)$ is the immediate successor, Trichotomy of the order on $N$ ). Transitivity of $\leq$ follows from the definition and associativity: $m + j = n$ and $n + i = p$ give $m + (j + i) = p$ .

[L6]

The canonical natural (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field): $ι (0) = 0_{R}$ and $ι (σ (n)) = ι (n) + 1_{R}$ ; $ι (n)$ is also written $n \cdot 1_{R}$ .

[L7]

For $n \geq 1$ , with $n \cdot 1_{F}$ defined by $1 \cdot 1_{F} = 1_{F}$ and $(n + 1) \cdot 1_{F} = n \cdot 1_{F} + 1_{F}$ : $n \cdot 1_{F} > 0$ , and $(m + n) \cdot 1_{F} = m \cdot 1_{F} + n \cdot 1_{F}$ and $(m n) \cdot 1_{F} = (m \cdot 1_{F}) (n \cdot 1_{F})$ for all $m, n \geq 1$ (Canonical naturals are positive and strictly increasing). These identities are asserted for $m, n \geq 1$ only; the cases with a zero argument are checked separately below.

[L8]

In a field, $0 \cdot x = 0$ (Multiplication by zero: $0 \cdot a = 0$ ); and $R$ is an ordered field, so its addition and multiplication are associative and commutative with identities $0$ and $1$ , and its order is total and compatible with addition (Field, Ordered field).

[L9]

Cancellation in $N$ : $m \cdot k = n \cdot k$ with $k \neq 0$ implies $m = n$ (Cancellation for multiplication by a nonzero factor); and $σ (n) \neq 0$ , so $1 \neq 0$ (The von Neumann naturals form a Peano system).

Proof

technique · induction

1.1given

Every clause is proved by induction on the upper index, using only the recursion clauses [L2], [L3] and the arithmetic [L4], [L8]; the inductions are written out one clause at a time.

1.2L1L6L7

The two notations agree: $ι (n) = n \cdot 1_{R}$ for every $n \geq 1$ . At $n = 1$ , $ι (1) = ι (σ (0)) = ι (0) + 1_{R} = 1_{R} = 1 \cdot 1_{R}$ ; and the successor clauses of the two recursions coincide, $ι (σ (n)) = ι (n) + 1_{R}$ and $(n + 1) \cdot 1_{R} = n \cdot 1_{R} + 1_{R}$ . So the two agree at every $n \geq 1$ by induction, and [L7] may be read as a statement about $ι$ .

1.3baseL2L4

Clause 1 at $n = 0$ : both sides are the empty sum, $0 = 0 + 0$ .

1.4ih

Clause 1, inductive hypothesis: assume $\sum_{k < n} (a_{k} + b_{k}) = \sum_{k < n} a_{k} + \sum_{k < n} b_{k}$ for a fixed $n$ and all lists $a, b$ .

1.5L1L2L4

Clause 2, by induction on $n$ . At $n = 0$ both sides are $0$ , since $0 \cdot c = 0$ . If $\sum_{k < n} c = n \cdot c$ , then $\sum_{k < σ (n)} c = \sum_{k < n} c + c = n \cdot c + c = σ (n) \cdot c$ , the last equality being the successor law $σ (a) \cdot n = a \cdot n + n$ of [L4].

1.6L1L2L4L5

Clause 3, by induction on $d$ , with $m$ fixed and $n = m + d$ . At $d = 0$ we have $n = m$ and the second sum is empty, so the claim reads $\sum_{k < m} a_{k} = \sum_{k < m} a_{k} + 0$ . Assuming it at $d$ , and using $m + σ (d) = σ (m + d)$ , we get $\sum_{k < σ (n)} a_{k} = \sum_{k < n} a_{k} + a_{n} = (\sum_{k < m} a_{k} + \sum_{j < d} a_{m + j}) + a_{m + d} = \sum_{k < m} a_{k} + (\sum_{j < d} a_{m + j} + a_{m + d}) = \sum_{k < m} a_{k} + \sum_{j < σ (d)} a_{m + j}$ . The product form is the same argument with $+$ replaced by $\cdot$ and $0$ by $1$ .

1.7L1L2L4L9

Clause 5, by induction on $n$ . At $n = 0$ both sides are $1 = 1 \cdot 1$ ; and $\prod_{k < σ (n)} (a_{k} b_{k}) = (\prod_{k < n} (a_{k} b_{k})) a_{n} b_{n} = (\prod_{k < n} a_{k}) (\prod_{k < n} b_{k}) a_{n} b_{n} = (\prod_{k < σ (n)} a_{k}) (\prod_{k < σ (n)} b_{k})$ by associativity and commutativity. For the second assertion, note first that a product of two nonzero naturals is nonzero: if $x y = 0$ with $y \neq 0$ , then $x y = 0 \cdot y$ , so $x = 0$ by [L9]. Now induct: $\prod_{k < 0} a_{k} = 1 \neq 0$ by [L9], and $\prod_{k < σ (n)} a_{k} = (\prod_{k < n} a_{k}) a_{n}$ is a product of two nonzero naturals.

2.1step 1.4L2L4

Clause 1, inductive step. Using [L2] twice and the associativity and commutativity of addition, $\sum_{k < σ (n)} (a_{k} + b_{k}) = \sum_{k < n} (a_{k} + b_{k}) + (a_{n} + b_{n}) = (\sum_{k < n} a_{k} + \sum_{k < n} b_{k}) + (a_{n} + b_{n}) = (\sum_{k < n} a_{k} + a_{n}) + (\sum_{k < n} b_{k} + b_{n}) = \sum_{k < σ (n)} a_{k} + \sum_{k < σ (n)} b_{k}$ , where the inductive hypothesis of step 1.4 was used at the second equality.

2.2step 1.2L4L7L8

Clause 0. First $ι (1) = 1$ , computed in step 1.2. For $m, n \geq 1$ the two identities are [L7], read through step 1.2. If $n = 0$ then $ι (m + 0) = ι (m) = ι (m) + 0 = ι (m) + ι (0)$ , and $ι (m \cdot 0) = ι (0) = 0 = ι (m) \cdot 0 = ι (m) ι (0)$ by [L8]; the case $m = 0$ follows from these by the commutativity of addition and multiplication in $N$ and in $R$ . So both identities hold for all $m, n \in N$ .

2.3step 1.6L2L4L5

Clause 4. Monotonicity is an induction: at $n = 0$ both sums are $0$ ; and if $\sum_{k < n} a_{k} \leq \sum_{k < n} b_{k}$ and $a_{n} \leq b_{n}$ , then $\sum_{k < n} a_{k} + a_{n} \leq \sum_{k < n} b_{k} + a_{n} \leq \sum_{k < n} b_{k} + b_{n}$ by [L5], so $\sum_{k < σ (n)} a_{k} \leq \sum_{k < σ (n)} b_{k}$ by transitivity. For the second assertion let $j < n$ , so $1 \leq n - j$ ; splitting at $j$ and then splitting the tail at $1$ , and using $\sum_{i < 1} a_{j + i} = 0 + a_{j} = a_{j}$ , gives $\sum_{k < n} a_{k} = (\sum_{k < j} a_{k} + a_{j}) + R$ for some $R \in N$ , and $a_{j} \leq (\sum_{k < j} a_{k} + a_{j}) + R$ because $a_{j}$ plus something equals it.

3.1step 1.3step 2.1L1

Clause 1 holds for every $n$ , by step 1.3 and step 2.1 together with induction.

3.2step 2.2L5L7L8

Clause 7. If $m < n$ put $d = n - m$ , so $m + d = n$ and $d \neq 0$ , hence $d \geq 1$ and $ι (d) > 0$ by [L7]; then $ι (n) = ι (m) + ι (d) > ι (m)$ by step 2.2. Conversely, if $ι (m) < ι (n)$ then $m = n$ and $n < m$ are both excluded, the first because the order of $R$ is irreflexive and the second by what was just proved, so $m < n$ by trichotomy in $N$ . The statement about equality follows by trichotomy on both sides.

3.3step 2.2L1L2L3L6

Clause 6, by induction on $n$ . At $n = 0$ , $ι (\sum_{k < 0} a_{k}) = ι (0) = 0 = \sum_{k < 0}^{R} ι (a_{k})$ . Assuming the identity at $n$ , $ι (\sum_{k < σ (n)} a_{k}) = ι (\sum_{k < n} a_{k} + a_{n}) = ι (\sum_{k < n} a_{k}) + ι (a_{n}) = \sum_{k < n}^{R} ι (a_{k}) + ι (a_{n}) = \sum_{k < σ (n)}^{R} ι (a_{k})$ , the second equality by step 2.2. The product form is the same induction, starting from $ι (1) = 1$ and using multiplicativity.

4.1step 1.5step 1.6step 1.7step 2.3step 3.1step 3.2step 3.3discharge-induction∎

Clause 0 is step 2.2, clause 1 is step 3.1, clause 2 is step 1.5, clause 3 is step 1.6, clause 4 is step 2.3, clause 5 is step 1.7, clause 6 is step 3.3 and clause 7 is step 3.2.

Remarks

Why the zero cases are done by hand. Canonical naturals are positive and strictly increasing states $(m + n) \cdot 1_{F} = m \cdot 1_{F} + n \cdot 1_{F}$ and $(m n) \cdot 1_{F} = (m \cdot 1_{F}) (n \cdot 1_{F})$ for $m, n \geq 1$ only, because the notation $n \cdot 1_{F}$ is introduced there by a recursion that starts at $1$ . Every count on this page can be $0$ , so the two one-line checks at $0$ in step 2.2 are not pedantry: without them clause 0 would be a citation to a statement that was not made.
The real-valued laws are the same list. Laws of finite sums and finite products proves additivity, scaling, splitting, monotonicity, telescoping and the product laws for sums of reals. The clauses above are their $N$ -valued counterparts, proved from the same recursion, and clause 6 is what ties the two lists together. Neither list contains a permutation-invariance clause; that is proved separately in the next item, and it is what the sum over a finite index set needs.
What clause 7 buys. Because $ι$ is injective, a proof may cross into $R$ , use subtraction or division there, and come back: if $ι (x) = ι (y)$ with $x, y \in N$ then $x = y$ . The binomial theorem below lives in $R$ for exactly this reason, while every coefficient in it is a count.

Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Laws of finite sums and products in NN, and ι(∑k<nak)=∑k<nι(ak)ι(∑k<n​ak​)=∑k<n​ι(ak​)

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$