User:Prime.mover/Proof Structures

Ordinary proofs
...etc.

$(1)$ implies $(2)$
Let $...$ be a [definiend] by definition 1.

Then by definition:
 * [Definition 1 of definiend]



Thus $...$ is a [definiend] by definition 2.

$(2)$ implies $(1)$
Let $...$ be a [definiend] by definition 2.

Then by definition:
 * [Definition 2 of definiend]



Thus $...$ is a [definiend] by definition 1.

]

Integration by Parts
With a view to expressing the primitive in the form:
 * $\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

and let:

Then:

Partition Proofs
Testing each of the criteria for a partition as follows:

Each element in no more than one component
Thus each element of ... is in no more than one component of ... .

Each element in at least one component
Thus each element of ... is in at least one component of ... .

No component is empty
Thus no component of ... is empty.

Thus ... is a partition by definition.

Metric Space proofs
It is to be demonstrated that $d$ satisfies all the metric space axioms.

Proof of $M1$
So axiom $M1$ holds for $d$.

Proof of $M2$
So axiom $M2$ holds for $d$.

Proof of $M3$
So axiom $M3$ holds for $d$.

Proof of $M4$
So axiom $M4$ holds for $d$.

Thus $d$ satisfies all the metric space axioms and so is a metric.

Topology Proofs
Each of the open set axioms is examined in turn:

$O1$: Union of Open Sets
Let $\left \langle{U_i}\right \rangle_{i \mathop \in I}$ be an indexed family of open sets of $T$.

Let $\displaystyle V = \bigcup_{i \mathop \in I} U_i$ be the union of $\left \langle{U_i}\right \rangle_{i \mathop \in I}$.



Hence $V$ is open by definition.

$O2$: Intersection of Open Sets
Let $U$ and $V$ be open sets of $T$.



Hence $U \cap V$ is open by definition.

$O3$: Underlying Set


All the open set axioms are fulfilled, and the result follows.

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

Reflexivity
Thus ... is seen to be reflexive.

Symmetry
Thus ... is seen to be symmetric.

Transitivity
Thus ... is seen to be transitive.

... has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

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

Reflexivity
So ... has been shown to be reflexive.

Transitivity
So ... has been shown to be transitive.

Antisymmetry
So ... has been shown to be antisymmetric.

... has been shown to be reflexive, transitive and antisymmetric.

Hence by definition it is an ordering.

Strict Ordering Proofs
Checking in turn each of the criteria for a strict ordering:

Antireflexivity
So ... has been shown to be antireflexive.

Transitivity
So ... has been shown to be transitive.

Asymmetry
So ... has been shown to be asymmetric.

... has been shown to be antireflexive, transitive and asymmetric.

Hence by definition it is a strict ordering.

Group Proofs
Taking the group axioms in turn:

$G \, 0$: Closure
Thus $...$ and so $...$ is closed.

$G \, 1$: Associativity
Thus $...$ is associative.

$G \, 2$: Identity
Thus $...$ is the identity element of $...$.

$G \, 3$: Inverses
We have that $...$ is the identity element of $\struct {\R, \circ}$.

Thus every element of $...$ has an inverse $...$.

All the group axioms are thus seen to be fulfilled, and so $...$ is a group.

Group Action Proofs
The group action axioms are investigated in turn.

Let $g, h \in G$ and $s \in S$.

Thus:



demonstrating that group action axiom $GA\,1$ holds.

Then:

demonstrating that group action axiom $GA\,2$ holds.

The group action axioms are thus seen to be fulfilled, and so $*$ is a group action.

Ring Proofs
Taking the ring axioms in turn:

Proof by Mathematical Induction
The proof will proceed by the Principle of Mathematical Induction on $\N$.

Let $S$ be the set defined as:
 * $S := \left\{ {n \in \N: ...}\right\}$

That is, $S$ is to be the set of all $n$ such that:

Basis for the Induction
We have that:

(proof that $1 \in S$)

So $1 \in S$.

This is our basis for the induction.

Induction Hypothesis
It is to be shown that, if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$.

This is the induction hypothesis:
 * $\text{expression for $k$}$

It is to be demonstrated that it follows that:
 * $\text{expression for $k + 1$}$

Induction Step
This is our induction step:

So $k \in S \implies k + 1 \in S$ and the result follows by the Principle of Mathematical Induction:


 * $\forall n \in \N: ...$

Tableau proofs
...etc.