User:Prime.mover/Proof Structures

Ordinary proofs

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

...etc.

Equivalence Proofs

Checking in turn each of the criteria for equivalence:

Reflexivity

So ... has been shown to be reflexive.

$\Box$

Symmetry

So ... has been shown to be symmetric.

$\Box$

Transitivity

So ... has been shown to be transitive.

$\Box$

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

Hence by definition it is an equivalence relation.

$\blacksquare$

Ordering Proofs

Checking in turn each of the criteria for an ordering:

Reflexivity

So ... has been shown to be reflexive.

$\Box$

Transitivity

So ... has been shown to be transitive.

$\Box$

Antisymmetry

So ... has been shown to be antisymmetric.

$\Box$

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

Hence by definition it is an ordering.

$\blacksquare$

Strict Ordering Proofs

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

Antireflexivity

So ... has been shown to be antireflexive.

$\Box$

Transitivity

So ... has been shown to be transitive.

$\Box$

Asymmetry

So ... has been shown to be asymmetric.

$\Box$

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

Hence by definition it is a strict ordering.

$\blacksquare$

Group Proofs

Taking the group axioms in turn:

G0: Closure

Thus ... and so ... is closed.

$\Box$

G1: Associativity

Thus ... is associative.

$\Box$

G2: Identity

Thus ... has an identity element.

$\Box$

G3: Inverses

Thus every element of ... has an inverse.

$\Box$

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

$\blacksquare$

Ring Proofs

Taking the ring axioms in turn:

Proof by Mathematical Induction

Proof by induction:

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

$proposition_n$

$P \left({1}\right)$ is true, as this just says $proposition_1$.

Basis for the Induction

$P \left({2}\right)$ 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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

Therefore:

$proposition_n$

Tableau proofs

Line		Pool	Formula	Rule	Depends upon	Notes
<line number>		<line numbers>	$\ldots{}$	link to ProofWiki entry	<line numbers>
<line number>		<line numbers>	$\ldots{}$	link to ProofWiki entry	<line numbers>

...etc.