logical negation of (a implies b)

Question

I know that the logical negation of $$\neg(a \rightarrow b)= a \wedge \neg b $$ I am not clear what that means in the following simple setting:

So its clear that $$x\geq 2 \to x^2\geq 4.$$ Now I can write the logical negation of $a\to b$ as $a \wedge \neg b$, but what does that intuitively mean?

Suppose I want to prove "$a \wedge \neg b$", what do i need to prove mathematically?

thnks

Answer 1

Say $a$ is $x\ge2$ and $b$ is $x^2\ge14$ (and formally an universal quantifier should be involved as in @JMoravitz comment, i.e. $\forall x, x\ge2\to x^2\ge14$ ). Pick $x=3$, then $a$ is true, but $b$ is false. In other words, $x\ge2$ does not imply that $x^2\ge14$. Formally, the negation here is $\exists x, (x\ge2 \land\neg x^2\ge14)$.

Formally, one may separate syntax from sematic (or form from meaning). Given any statements $a$ and $b$, the negation of the formula $a\to b$ is the formula $a\land\neg b$. If you want to prove that $a\land\neg b$ then you need to either use some previously proven formulas, or axioms (accepted without proof), or to interpret $a$ and $b$ in some known model (as for the reals above), giving each of $a$ and $b$ meaning and truth values.

Answer 2

a is $x\geq 2$ and $\neg b$ is $x^2\lt 4$.

So, the intuitive meaning of $a \wedge \neg b$ is that both of these cannot happen at the same time, $x\geq 2$ with $x^2\lt 4$ have no common elements in their solution sets.

To prove $a \wedge \neg b$, you should show that all elements in the domain of x satisfy both $x\geq 2$ and $x^2\lt 4$. However, this is impossible and that is why the negation of a true logical statement is false.

Answer 3

Our statement is $\neg (a\to b)$

This reads: "It is false that $a$ (materially)implies $b$".

Recall that a material implication is falsified only when the antecedant is false and the consequent is true.

So our statement must be infering that "$a$ is true and $b$ is false."

Which is written $a\wedge\neg b$.

And we can argue vice versa, so the statements are equivalent.

That is all.

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Now the negation of $\bbox[lemonchiffon]{x\geq 2\to x^2\geq 4}$ is $\bbox[lemonchiffon]{x\geq 2\wedge x^2< 4}$.   If the latter is false the former will be true (and vice versa).

For example, when $x=3$ then $\bbox[lemonchiffon]{3\geq 2\to 3^2\geq 4}$ is true because $\bbox[lemonchiffon]{3\geq 2\wedge 3^2< 4}$ is false.

Another example, when $x=1$ then $\bbox[lemonchiffon]{1\geq 2\to 1^2\geq 4}$ is true (despite seeming absurd) because $\bbox[lemonchiffon]{1\geq 2\wedge 1^2< 4}$ is false.

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Often when we write something like $x\geq 2\to x^2\geq 4$ we implicitly mean that the statement holds universally (for all $x$).   That is $\forall x~(x\geq 2\to x^2\geq 4)$.

The negation of this quantified statement is $\exists x~(x\geq 2\wedge x^2<4)$.   Since there is no real witness to this exitential, then the universal is infered to be true.