Non-symmetrical relation: a mistake while asserting "not implies" with increasing levels of "strength"?

Question

In this section of video from a book group I'm following we consider the lack of symmetry of the relation $\lt$ on $\mathbb{R}$. The condition for symmetry, which does not hold, would look like this

$$ \forall a,b \in \mathbb{R}, a < b \implies b < a \tag{1}\label{eq1} $$

The video then offers three ways that this condition is negated, in increasing strength of the negation:

$$\begin{align} \exists a,b \in \mathbb{R}&, a\lt b \,\,\,\,\not\!\!\!\!\implies b\lt a \tag{2}\label{eq2} \\ \forall a,b \in \mathbb{R}&, a\lt b \,\,\,\,\not\!\!\!\!\implies b\lt a \tag{3}\label{eq3} \\ \forall a,b \in \mathbb{R}&, a\lt b \implies \lnot(b \lt a) \tag{4}\label{eq4} \end{align}$$

(The video explains that the notation $P \,\,\,\,\not\!\!\!\!\implies Q$ is the same as $\lnot(P\implies Q)$.)

I can immediately understand that $\eqref{eq2}$ is a simple negation of $\eqref{eq1}$. I can then understand how each statement that follows is a stronger negation: the existential quantifier is replaced by the universal in $\eqref{eq3}$, and then not implies is replaced by implies not in $\eqref{eq4}$.

However, is $\eqref{eq3}$ a true statement? The video suggests that $\eqref{eq1}$ is false, while $\eqref{eq2}$, $\eqref{eq3}$ and $\eqref{eq4}$ are true. I agree except that I've come to think that $\eqref{eq3}$ is false. I'm not confident that my reasoning (see below) is correct and would appreciate either some reassurance or a correction.

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How I arrived at this question

I was immediately interested in whether one of $\eqref{eq3}$ or $\eqref{eq4}$ was equivalent to saying that $\lt$ is antisymmetric. After some reading on Wikipedia, I discovered that $\eqref{eq4}$ is a statement of asymmetry (a stronger condition), and so then I wondered if that leaves $\eqref{eq3}$ as a statement of the weaker antisymmetry. When I failed to understand how $\eqref{eq3}$ might be related to the definition of antisymmetry given in Wikipedia, I realised I didn't understand not implies ($\,\,\,\,\not\!\!\!\!\implies$) very well, so I searched for information on negation of implication.

From reading What is the negation of the implication statement I learned I could use

$$\lnot(P\implies Q) \iff P\land\lnot Q$$

to modify $\eqref{eq2}$ and $\eqref{eq3}$ as follows:

$$\begin{align} \exists a,b \in \mathbb{R}&, a\lt b \land \lnot(b\lt a) \tag{2*}\label{eq2'} \\ \forall a,b \in \mathbb{R}&, a\lt b \land \lnot(b\lt a) \tag{3*}\label{eq3'} \end{align}$$

While $\eqref{eq2'}$ seems fine, I think $\eqref{eq3'}$ is false. Surely $a$ is not less than $b$ for all $a$ and $b$. Have I made a mistake, or is statement $\eqref{eq3}$ actually false?

Answer 1

The video then offers three ways that this condition is negated, in increasing strength of the negation:

\begin{align} \exists a,b \in \mathbb{R}&, a\lt b\,\,\,\,\not\!\!\!\!\implies b\lt a \tag{2}\\ \forall a,b\in \mathbb{R}&, a\lt b \,\,\,\,\not\!\!\!\!\implies b\lt a\tag{3} \\ \forall a,b \in \mathbb{R}&, a\lt b \implies\lnot(b \lt a) \tag{4}\end{align}

Since a sentence and its negation must have opposite truth values regardless of context (that is, they logically contradict each other), that claim/concept of increasing strength of negation is dubious. For example, consider the sentences $P,Q,R$ such that $R$ is a logical consequence of $Q$ but not vice versa and $Q$ is a logical consequence of $P$ but not vice versa:

• suppose that $\lnot Q$ is true; then $P$ and $Q$ are both false but $R$ can be either true or false
• suppose that $\lnot Q$ is false, then $Q$ and $R$ are both true but $P$ can be either true or false

hence, only $Q$—but neither $P$ nor $R$—definitely has the opposite truth value of $\lnot Q.$

Among our sentences $(2), (3)$ and $(4),$ the only logical-entailment relationships (out of six possible permutations) are: $$(3)\models(2)\\(3)\models(4).$$

The video suggests that $(1)$ is false, while $(2)$, $(3)$ and $(4)$ are true.

Then the video is wrong: while both $(2)$ and $(4)$ turn out to be mathematically true, the counterexample $(a,b)=(0,0)$ shows that $(3)$ is mathematically false.

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The video's notion of negation strength may be a misplacement of the idea that non-implication (similarly: implication) can be asserted at various levels of logical strength.

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I think $(3)$ is false. Surely $a$ is not less than $b$ for all $a$ and $b$.

Exactly.

Have I properly shown that (3*) is false, that (3*) follows from (3) and that therefore (3) is false?

Correction: the conclusion in your last clause results from (3*) and (3) being equivalent rather than from (3*) merely following from (3).

P.S. I usually understand $$Px{\kern.6em\not\kern-.6em\implies} Qx$$ as a shorthand for the existential quantification $\;\exists x\:(Px\land \lnot Qx),\,$ so $$\forall a{,}b{\in} \mathbb{R} \;(a\lt b{\kern.6em\not\kern-.6em\implies }b\lt a)\tag3$$ is quite unusual to parse.