User:Prime.mover

I'm on the local email as prime.mover@proofwiki.org so drop us a line when you want.

You might also catch me asking (but usually answering) questions on http://www.mathhelpforum.com/math-help/ which I haunt when ProofWiki's down.

what's THIS for ...!
Useful constructs for anyone to cut and paste.

For example:

Let $$\mathcal{P} \left({S}\right)$$ be the power set of the set $$S$$.

Let $$\left({S, \circ}\right)$$ be an algebraic structure or a semigroup or a group.

Let $$\left({R, +, \circ}\right)$$ be a ring whose zero is $$0_R$$ and whose unity is $$1_R$$.

Let $$J$$ be an ideal of $$R$$.

Let $$\left({R / J, +, \circ}\right)$$ be the quotient ring defined by $$J$$.

Let $$\left({D, +, \circ}\right)$$ be an integral domain whose zero is $$0_D$$ and whose unity is $$1_D$$.

Let $$\left({K, +, \circ}\right)$$ be a division ring whose zero is $$0_K$$ and whose unity is $$1_K$$.

Let $$\left({F, +, \circ}\right)$$ be a field whose zero is $$0_F$$ and whose unity is $$1_F$$.

Let $$\left({K, +, \circ}\right)$$ be a quotient field of an integral domain $$\left({D, +, \circ}\right)$$.

Let $$\left({D, +, \circ; \le}\right)$$ be a totally ordered integral domain whose zero is $$0_D$$ and whose unity is $$1_D$$.

Let $$\left({S; \preceq}\right)$$ be a totally ordered set.

Let $$\left({S, \circ; \preceq}\right)$$ be an ordered structure.

Let $$\left({S, \circ; \preceq}\right)$$ be a naturally ordered semigroup.

Let $$\left({S, \circ, \ast; \preceq}\right)$$ be a Naturally Ordered Semigroup with Product.

$$\left[{m \,. \, . \, n}\right]$$ is the closed interval between $m$ and $n$.

$$\mathbf {Define:} \ fred \ \stackrel {\mathbf {def}} {==} \ bert$$

$$\mathbb{N}$$, $$\mathbb{N}^*$$, $$\mathbb{N}_k$$, $$\mathbb{N}^*_k$$

$$\mathbb{Z}$$, $$\mathbb{Z}^*$$, $$\mathbb{Z}_+$$, $$\mathbb{Z}^*_+$$,

$$\mathbb{Q}$$

$$\mathbb{R}$$

$$\mathbb{C}$$

The cardinality of a set $$S$$ is written $$\left|{S}\right|$$.

Let $$\left \langle {s_k} \right \rangle_{k \in A}$$ be a sequence in $S$.

Let $$\gcd \left\{{a, b}\right\}$$ be the Greatest Common Divisor of $$a$$ and $$b$$.

Let $$\mathrm{lcm} \left\{{a, b}\right\}$$ be the Lowest Common Multiple of $$a$$ and $$b$$.

Let $$\left|{a}\right|$$ be the absolute value of $$a$$.

$$a \equiv b \left({\bmod\, m}\right)$$: "$$a$$ is congruent to $$b$$ modulo $$m$$."

$$\left[\left[{a}\right]\right]_m$$ is the congruence class of $$a$$ (modulo $$m$$).

Ordinary proofs
...etc.

Equivalence Proofs
Checking in turn each of the critera for equivalence:

Ordering Proofs
Checking in turn each of the critera for an ordering:

Group Proofs
Taking the group axioms in turn:

Ring Proofs
Taking the ring axioms in turn:

Proof by Mathematical Induction
Proof by induction:

For all $$n \in \mathbb{N}^*$$, let $$P \left({n}\right)$$ be the proposition $$proposition_n$$.


 * $$P(1)$$ is true, as this just says $$proposition_1$$.

Basis for the Induction

 * $$P(2)$$ is the case $$proposition_2$$, which has been proved above. This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:

$$proposition_k$$.

Then we need to show:

$$proposition_{k+1}$$.

Induction Step
This is our induction step:

So $$P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$proposition_n$$.

Tableau proofs
...etc.

Logical Axiom references
These are for tableau proofs:


 * Declaration of a Proposition: P


 * Rule of Assumption: A


 * Rule of Conjunction: $\land \mathcal{I}$


 * Rule of Simplification: $\land \mathcal{E}_1$ or $\land \mathcal{E}_2$


 * Rule of Addition: $\lor \mathcal{I}_1$ or $\lor \mathcal{I}_2$


 * Rule of Or-Elimination: $\lor \mathcal{E}$


 * Modus Ponendo Ponens: $\Longrightarrow \mathcal{E}$


 * Rule of Implication: $\Longrightarrow \mathcal{I}$


 * Rule of Not-Elimination: $\lnot \mathcal{E}$


 * Rule of Proof by Contradiction: $\lnot \mathcal{I}$


 * Rule of Bottom-Elimination: $\bot \mathcal{E}$


 * Law of the Excluded Middle: $\textrm{LEM}$


 * Double Negation Introduction: $\lnot \lnot \mathcal{I}$


 * Double Negation Elimination: $\lnot \lnot \mathcal{E}$

Barnstars
The tireless contributor barnstar for all the long hours you have spent adding to the site. Thank you and congratulations!