Justifying lines in proof: why $\neg (a=b) \Leftrightarrow a<b \vee a>b$

Question

Two questions similar questions:

$(1)$ I am performing a proof and wondering how I justify the following step...

$\{ (a,b)\in \mathbb{R}^2 | a=b \vee \neg(a>b) \} = \{ (a,b)\in \mathbb{R}^2 | a=b \vee a \le b \} $

Can I say something like by Disjunctive syllogism with trichotomy law?

$(2)$ I am also trying to justify the first step below

$\{ (a,b)\in \mathbb{R}^2 | \neg (a \neq b) \wedge a>b \} = \{ (a,b)\in \mathbb{R}^2 | (a<b \vee a>b) \wedge a>b \} $ ...???

$=\{ (a,b)\in \mathbb{R}^2 | a>b \}$ by the aborption law

Thanks for your insights!

Answer 1

For all x in R, (x < 0 or x = 0 or 0 < x ).

Assume a,b in R.
Thus a - b < 0 or a - b = 0 or 0 < a - b.
Whereupon a < b or a = b or b < a.

So if a /= b, then a < b or b < a.
Conversely, if a < b or b < a, then a /= b.
Otherwise if a = b, then a < a, a no, no.

The use of sets as you are doing is making a
difficult and clumsy proof. Use logic instead.

Answer 2

Your method of proving it is making it difficult and clumsy as William already stated. Below you can find another way to prove it.

We first prove that:

$$ a \neq b\Rightarrow a < b \lor a > b $$

You can use contraposition. So we prove that $$a \leq b \land a \geq b \Rightarrow a = b$$

We know that $a \leq b$ and $a \geq b$ so $a \leq b \leq a$, this means that $a=a$, thus $a=b$ must also hold. Because of the use of contraposition, we can conlude that $a \neq b\Rightarrow a < b \lor a > b $.

Now try to prove

$$a < b \lor a > b \Rightarrow a \neq b.$$

Answer 3

First one:

$\{ (a,b)\in \mathbb{R}^2 | a=b \vee \neg(a>b) \} = \{ (a,b)\in \mathbb{R}^2 | a=b \vee a \le b \} $

Can I say something like by Disjunctive syllogism with trichotomy law?

Trichotomy is

$$a=b \lor a< b \lor a < b$$, and so if $\neg(a>b)$, then (by Disjunctive Syllogism) $a = b \lor a < b$ ... which is equivalent to (typically, the definition of) $a \leq b$. And that, since $a=b$ implies $a \leq b$ is equivalent to $a = b \lor a \leq b$.

However, for going the other way around, you need Asymmetry of $<$:

$$a<b \to \neg (b < a)$$

so that given $a = b \lor a < b$, you get $a = b \lor \neg (b < a)$.

And technically, you then also need the connection between $<$ and $>$, i.e.

$$a < b \leftrightarrow b > a$$

to get to $a = b \lor \neg (a > b)$

Second one:

$\{ (a,b)\in \mathbb{R}^2 | \neg (a \neq b) \wedge a>b \} = \{ (a,b)\in \mathbb{R}^2 | (a<b \vee a>b) \wedge a>b \} $ ...???

You have two negations in $\neg a \not = b$. I think you meant only one:

$\{ (a,b)\in \mathbb{R}^2 | \neg (a = b) \wedge a>b \} = \{ (a,b)\in \mathbb{R}^2 | (a<b \vee a>b) \wedge a>b \} $ ...???

Again, from left to right is Trichotomy and Disjunctive Syllogism. From right to left you need Irreflexivity:

$$\neg (a < a)$$

because given $(a < b \lor b < a)$ we can then get $\neg(a=b)$ by Proof by contradiction: If $a=b$, then $(a < b \lor b < a)$ becomes $(a < a \lor b < b)$, but both disjuncts contradict Irreflexivity.