User:Lord Farin/Proof Structures

= Induction =

Proof by Mathematical Induction (type 1 - 'Normal' Induction)
The proof proceeds by induction on $n$, the describe the integer $n$.

Basis for the Induction
The case $n = n_0$ is verified as follows:


 * Reasoning for $n_0$

This is the basis for the induction.

Induction Hypothesis
Fix $n \in \N$ with $n \ge n_0$.

Assume to-be-proved property holds for $n$.

This is our induction hypothesis.

Induction Step
This is our induction step:

''Insert reasoning. Whenever the induction hypothesis is used, write down explicitly that this is done.''

When desired, use the reference induction hypothesis.

We conclude to-be-proved property holds for $n+1$.

The result follows by the Principle of Mathematical Induction.

Proof by Mathematical Induction (type 2 - Complete Induction)
The proof proceeds by induction on $n$, the describe the integer $n$.

Basis for the Induction
The case $n = n_0$ is verified as follows:


 * Reasoning for $n_0$

This is the basis for the induction.

Induction Hypothesis
Fix $n \in \N$ with $n \ge n_0$.

Assume to-be-proved property holds for all $k$ such that $n_0 \le k \le n$.

This is our induction hypothesis.

Induction Step
This is our induction step:

''Insert reasoning. Whenever the induction hypothesis is used, write down explicitly that this is done.''

When desired, use the reference induction hypothesis.

We conclude to-be-proved property holds for $n+1$.

The result follows by the Second Principle of Mathematical Induction.

Notes to Induction Proofs
The above are only guidelines. In most cases, a different formulation is natural. Just use this as a template, so as to structure all induction proofs uniformly. Some proofs require induction to multiple variables. I haven't encountered such proofs on PW; when I do, I will put up a proof template.

= Set Equality =

Let $A$ be describe the set $A$.

Let $B$ be describe the set $B$.

Then we have $A = B$.

Proof of Set Equality (type 1 - Direct)
It suffices to prove $A \subset B$ and $B \subset A$.

$A$ is contained in $B$
Suppose that $x \in A$.

Insert reasoning

Hence $x \in B$. It follows that $A \subset B$.

$B$ is contained in $A$
Suppose now that $x \in B$.

Insert reasoning

Hence $x \in A$. It follows that $B \subset A$.

Therefore, we have established that $x \in B \iff x \in A$.

From the definition of set equality, we conclude that $A = B$.

Proof of Set Equality (type 2 - Indirect)
It suffices to prove $A \subset B$ and $\complement \left({A}\right) \subset \complement \left({B}\right)$.

$A$ is contained in $B$
Suppose that $x \in A$.

Insert reasoning

Hence $x \in B$. It follows that $A \subset B$.

$\complement \left({A}\right)$ is contained in $\complement \left({B}\right)$
Suppose now that $x \notin A$.

Insert reasoning

Hence $x \notin B$. It follows that $\complement \left({A}\right) \subset \complement \left({B}\right)$.

Therefore, we have established that $x \in B \iff x \in A$.

From the definition of set equality, we conclude that $A = B$.

Notes to Set Equality Proofs
I am pretty sure no convention on this type of proof exists. Encountering some of these scattered around, I figured I'd give it a shot.

= General Notes =

For further proof templates, cf. User:Prime.mover/Proof Structures.