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Suppose $A$ does not imply $B$ and that $B$ implies $C.$ Does it follow that $A$ does not imply $C$? I think this is false.

I was doing an Analysis exercise and realized I think I had some bungled logic. Here are the statements I was working with: $$\text{If a function is differentiable then it is not necessarily continuously differentiable.} \\\text{If a function is continuously differentiable then it is continuous.}$$ Putting them together: $$\text{If a function is differentiable then it is not necessarily continuous.}$$ Which is a false statement.

ryang
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yoshi
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    It's clearly true that if a function is (continuously or not) differentiable then it is continous (you can think of these statements as $A,B,C$). So $A\implies C$ and $B\implies C$ but a differentiable function doesn't mean continuously differentiable function, so $A$ does not imply $B$. I think this is a clear enough counterexample. – user160738 Nov 27 '18 at 14:10

3 Answers3

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Suppose $A$ does not imply $B$ and $ B \implies C$. Does it follow that $A$ does not imply $C$? I think this is false.

Sometimes it is false.

$\neg(A \implies B) \land (B\implies C) \implies \neg (A\implies C)$ is a contingency, i.e. it is sometimes true and sometimes false. It is true when, for example, $A, B$ and $C$ are all true. It is false when $A$ and $C$ are both true and $B$ is false. See the truth table.

Dan Christensen
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Suppose $A$ does not imply $B$ and $ B \implies C$. Does it follow that $A$ does not imply $C$? I think this is false.

Correct, that is false.

For example:

  • $A$ = "is greater than $1$"
  • $B$ = "is greater than $2$"
  • $C$ = "is greater than $0$"

then $B \implies C$ is true but $A \implies B$ does not hold; however $A \implies C$ is clearly true.

In this example, you have $B \subset A \subset C$ and similarly, you have: $$\mbox{"$f$ is continuously differentiable"}\subset \mbox{"$f$ is differentiable"} \subset \mbox{"$f$ is continuous"}$$

You could draw (Venn-)diagrams to illustrate this more clearly in a graphical way.

StackTD
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0

Suppose $A$ does not imply $B$ and that $B$ implies $C.$ Does it follow that $A$ does not imply $C$? I think this is false.

The argument $$\Big(\lnot (A\to B)\land(B\to C)\Big)\to\lnot(A\to C)\tag1$$ is not valid, because, as explained by Dan, it is a logical contingency, because its truth value varies according to what $A,B$ and $C$ stand for.

I was doing an Analysis exercise: $$\text{If a function is differentiable then it is not necessarily continuously differentiable.} \\\text{If a function is continuously differentiable then it is continuous.}$$ Putting them together: $$\text{If a function is differentiable then it is not necessarily continuous.}$$

However, sentence $(1)$ does not actually capture your Analysis argument, which is $$\Big(\lnot\forall f\: \big(D(f)\to CD(f)\big)\land\forall f\:\big(CD(f)\to C(f)\big)\Big)\to\lnot\forall f\: \big(D(f)\to C(f)\big)$$ instead, which is logically equivalent to $$\color{red}{∀f\,∃g\,∃h\,}\bigg(\Big(¬\big(D(\color\red f)→P(\color\red f)\big)∧\big(P(\color\red g)→C(\color\red g)\big)\Big)→¬\big(D(\color\red h)→C(\color\red h)\big)\bigg);$$ this argument is satisfiable but again invalid.

ryang
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