Counterexample to "$A \to B, A \to C$, therefore $B \to C$" 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...)
 A: \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}
A: $$x = -1 \implies x^4 = 1$$
$$x = -1 \implies x^3 = -1$$
But
$$x^4 = 1\not\Rightarrow x^3 = -1$$
because $x^3$ could be equals to $1$.
A: $A$: "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.
A: 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
=====A: is ice cream
B: starts with the letter I
C: is made from milk
Therefore Insects are made out of milk.
=====
The possibilities are endless.
