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In previous posts we defined porpositions, theorems, corollaries, among others. To demonstrate theorems, it means that the given proposition is a tautology, in which we obtain through the truth tables. Below, we present some methods for demonstrating results.

Deductive method

The deductive method is obtained through the use of definitions and other pre-established results. For this, we use the implications $(\Rightarrow)$ and logical bi-implications $(\Leftrightarrow)$.

Example 1: With the following statements, we will show whether it is a theorem or not. For this, the following hypotheses are:

• $H_1$: time is money.
• $H_2$: bum has a lot of time.

And the thesis:

• $T$: Tramp has a lot of money.

Is the proposition $H_1 \wedge H_2 \Rightarrow T$ a theorem?

Let’s suppose:

• $p$: “have time”.
• $q$: “have money”.
• $r$: “being a bum”.

With:

• $H_1$: $p\wedge q$ (having time and having money).
• $H_2$: $q\rightarrow r$ (having money implies being a bum).
• $T$: $r\rightarrow q$ (being a bum implies having money).

With this, we can write the proposition:

Hypothesis 1 and hypothesis 2 imply Thesis.

How is the tautology valid:

(having time implies having money) and (being a bum implies having time) implies (being a bum implies having money).

Therefore, the given proposition is a theorem.

From a practical point of view this result may not make sense, but depending on the statements used, we can reach an absurdity.

Example 2: With the following statements, we will show whether it is a theorem or not. For this, the following hypotheses are:

• $H_1$: I think, therefore I am.
• $H_2$: stones don’t think.

And the thesis:

• $T$: stones do not exist.

Is the proposition $H_1 \wedge H_2 \Rightarrow T$ a theorem?

Let’s suppose:

• $p$: “think”.
• $q$: “exist”.

With:

• $H_1$: $p\rightarrow q$ (thinking implies existing).
• $H_2$: $\sim p$ (don’t think).
• $T$: $\sim q$ (does not exist).

With this, we can write the proposition:

Hypothesis 1 and hypothesis 2 imply Thesis.

As it is not a tautology:

(thinking implies existing) and (not thinking) implies (not existing).

Therefore, the given proposition is not a theorem.

This last result is not even a result, as it is not even a theorem. With this, absurdity is natural.

Demonstration Methods

In this section, we will show how to prove a theorem $p\Rightarrow q$. Most income statements fall into one of the following three methods:

• Direct Method.
• Contraposition method.
• Method by Contradiction (Reduction ad absurdum).

Direct Method

This method is obtained from the deductive method, with the assumption that proposition $p$ is true and the consequence, proposition $q$ is a logical equivalence. Therefore, the objective is to deduce that $q$ is true.

Example 3: If $x$ is an even integer, then $x^2$ is an even integer.

Demonstration:

To construct this result, we assume the proposition $p$ is true, that is, $x$ is an even integer.

As a consequence, we have that $x$ is divisible by 2 (see post on even numbers).

By definition, $x$ is a multiple of 2.

Therefore, there exists $n\in \mathbb{Z}$ such that $x = 2n$.

When we square both sides of the equality, we have:

Therefore, $x^2$ is an even integer.

$\square$

Contraposition method

The contraposition method is obtained with logical equivalence:

More specifically, to demonstrate the theorem:

Let us simply demonstrate using the direct method:

Example 4: If $x$ is an integer such that $x^2$ is odd, then $x$ is an odd integer.

Demonstration:

Let $p$ and $q$ be the propositions:

• $p$: $x^2$ is an odd integer.
• $q$: $x$ is an odd integer.

and we want to prove the implication of $p \Rightarrow q$. In this case, the direct method does not apply, because when we assume that $x^2$ is an odd integer, there is $n\in \mathbb{Z}$ such that $x^2 = 2n+1$.

With this, we cannot deduce that there is an $m\in \mathbb{Z}$ such as $x = 2m+1$.

Instead, let’s use logical equivalence.

Let us suppose the proposition $\sim q: $ $x$ is not odd true, that is, $x$ is even.

From the example of the direct method, we have that this result implies that $\sim p:$ $x^2$ is not odd, that is, $x^2$ is even.

Therefore, $p \Rightarrow q$.

$\square$

Method by Contradiction

The contradiction method is obtained with logical equivalence:

More specifically, to demonstrate the theorem:

Let us simply demonstrate:

and we arrive at an absurdity, because if $p$ is true and we conclude that $\sim p$ is also true. In this case we arrive at a result

true, which is a contradiction.

Example 5: If $x$ is an integer such that $x^2$ is even, then $x$ is an even integer.

Demonstration:

Let $p$ and $q$ be the propositions:

• $p$: $x^2$ is an even integer.
• $q$: $x$ is an even integer.

and we want to prove the implication of $p \Rightarrow q$.

By the direct method, we assume that $(p \wedge \sim q)$ is true, that is,

• $x^2$ is even integer.
• $x$ is an odd integer.

From this, we get that $x=2n+1$ for some $n \in \mathbb{Z}$.

When we square both sides of the equality, we have:

for some $m\in \mathbb{Z}$

This implies $\Longrightarrow$ that $x^2$ is an odd integer ($\sim p$). Therefore, we reach an absurdity, as we assume $x^2$ even.

In this way, we summarize some techniques for demonstrating results.