Counterexample to "$A \to B, A \to C$, therefore $B \to C$"

Question

We have $A\to B$ and $A\to C$. I need counter-examples to: '$\therefore B\to C$'.

More formally, disprove:

$$ (A\to B)\land(A\to C)\to (B\to C)$$

I have $A$ is a blackbird, $B$ is 'is black', $C$ is 'is a bird', and so $B \not\to C$. Does anyone have a more numerical solution?

(alternatively, cases where this is true...)

Answer 1

\begin{align} \mathcal A & = \{5,6\} \\ \mathcal B & = \{1,2,3,4,5,6\} \\ \mathcal C & = \{3,4,5,6,7,8,9,10\} \\[12pt] A & = [x\in\mathcal A] \\ B & = [x\in\mathcal B] \\ C & = [x\in\mathcal C] \end{align}

Then $A\to B$ and $A\to C$ are true but $B\to C$ is false.

Second example: \begin{align} A & = [x\in\{1,2,3,4,\ldots\}] \\ B & = [x \text{ is rational}] \\ C & = [x>0] \end{align}

Answer 2

$A$: "I do not exist"
$B$: "I work for a living"
$C$: "I am a billionaire"

In other words, make $A$ false, $B$ true, $C$ false.

Answer 3

$$x = -1 \implies x^4 = 1$$ $$x = -1 \implies x^3 = -1$$

But

$$x^4 = 1\not\Rightarrow x^3 = -1$$

because $x^3$ could be equals to $1$.

Answer 4

Oh. You wanted numerical examples.

The number 2 is even, 2 is prime, therefore all even numbers are prime.

3 is a square root, 3 is rational. All square roots are rational.

Ooh here's a good one:

$20|x \implies 5|x$

$20|x \implies 4|x$

Ergo $5|x \implies 4|x$.

Etc. etc.

=====old answer ===

A: is a fish

B: lives in water

C: has gills.

Therefore sea otters have gills

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A: is ice cream

B: starts with the letter I

C: is made from milk

Therefore Insects are made out of milk.

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The possibilities are endless.