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Given a additively cancellative commutative semiring $R$, a term $e:R$ is left cancellative if for all $a:R$ and $b:R$, $e \cdot a = e \cdot b$ implies $a = b$.
A term $e:R$ is right cancellative if for all $a:R$ and $b:R$, $a \cdot e = b \cdot e$ implies $a = b$.
An term $e:R$ is cancellative if it is both left cancellative and right cancellative.
The multiplicative submonoid of cancellative elements in $R$ is the subset of all cancellative elements in $R$
A Euclidean semiring is a additively cancellative commutative semiring $R$ for which there exists a function $d \colon \mathrm{Can}(R) \to \mathbb{N}$ from the multiplicative submonoid of cancellative elements in $R$ to the natural numbers, often called a degree function, a function $(-)\div(-):R \times \mathrm{Can}(R) \to R$ called the division function, and a function $(-)\;\%\;(-):R \times \mathrm{Can}(R) \to R$ called the remainder function, such that for all $a \in R$ and $b \in \mathrm{Can}(R)$, $a = (a \div b) \cdot b + (a\;\%\; b)$ and either $a\;\%\; b = 0$ or $d(a\;\%\; b) \lt d(g)$.
Given a ring $R$, an element $x \in R$ is non-cancellative if: if there is an element $y \in \mathrm{Can}(R)$ with injection $i:\mathrm{Can}(R) \to R$ such that $i(y) = x$, then $0 = 1$. An element $x \in R$ is non-invertible if: if there is an element $y \in R^\times$ with injection $j:R^\times \to R$ such that $j(y) = x$, then $0 = 1$.
There is a significant difference between square roots and $n$-th roots. Square roots are the inverse operation of the diagonal $f(x, x)$ for any binary operation $f$, while $n$-th roots are inverse operations of the $n$-dimensional diagonals $g(x, x, \ldots, x)$ for $n$-ary operations, which we typically do not formally talk about in typical practice for rings, etc…
Given an Archimedean ordered integral domain $A$ with element $\gamma:A$, $p:0 \lt \gamma$, and $q:\gamma \lt 1$, a function $f:A \to A$ is pointwise smooth differentiable if it comes with a function sequence$D{D}^{(-)}(f):\mathbb{N}\to (A\to A)$ D(f):A D^{(-)}(f):\mathbb{N} \to A (A \to A) called the iterated derivative sequence and a function sequence of functions$M{M}^{(-)}:\mathbb{N}\to ({A}_{+}\to {A}_{+})$ M:A_+ M^{(-)}:\mathbb{N} \to A_+ (A_+ \to A_+) on in the positive elements of$A$ called the modulus moduli sequence of iterated differentiability , such that for every positive element$\epsilon:A_+$, for every element $h:A$ such that $0 \lt \max(h, -h) \lt M(\epsilon)$, and for every element $x:A$,
$D^{0}(f) = f$
for every natural number $n:\mathbb{N}$, for every positive element $\epsilon:A_+$, for every element $h:A$ such that $0 \lt \max(h, -h) \lt M^{n+1}(\epsilon)$, and for every element $x:A$,
Given an Archimedean ordered integral domain $A$ with element $\gamma:A$, $p:0 \lt \gamma$, and $q:\gamma \lt 1$, a function $f:A \to A$ is smooth if it comes with a sequence $D^{(-)}(f):\mathbb{N} \to (A \to A)$ called the iterated derivative sequence and a sequence of functions $M^{(-)}:\mathbb{N} \to (A_+ \to A_+)$ in the positive elements of $A$ called the moduli sequence of iterated differentiability, such that
$D^{0}(f) = f$
for every natural number $n:\mathbb{N}$, for every positive element $\epsilon:A_+$, for every element $h:A$ such that $0 \lt \max(h, -h) \lt M^{n+1}(\epsilon)$, and for every element $x:A$,
Given a commutative ring $R$, there is a commutative ring $A$ where $R$ is a subring of $A$, with a function $(-)\circ(-):A \times A \to A$ called composition, a term $x:A$ called the composition identity, a function $S_{(-)}:R \times A \to A$ called the shift, and a function $\partial:A \to A$ called the derivative such that
rules for composition:
rules for shifts:
rules for derivatives:
Symbolic representations of formal smooth functions on the entire domain.