# Talk:All Horses are the Same Colour

The induction step needs to use strong induction, no? Because $T_c$ is of size $k-1$? Or is that putting too much of the explanation of the resolution in the paradox... --GFauxPas (talk) 00:18, 3 July 2014 (UTC)

Actually, you have a point there. I wonder whether this particular point, where you need to take into account the size of $T_c$, is the crucial proof-breaker. Because you need to guarantee that the proof is true for sets of size $k$ and $k - 1$, that means you have to explicitly examine the case where $k = 1$ and $k = 2$, as the argument cannot be used for $k = 1$ as there is no concept of a set of size $-1$. I may be overthinking again ...--prime mover (talk) 07:29, 3 July 2014 (UTC)
I think you're on the money. How's this for an alternative resolution? --GFauxPas (talk) 12:42, 3 July 2014 (UTC)
...unless one says that this is really the same resolution as the first one, which I could definitely hear. --GFauxPas (talk) 13:09, 3 July 2014 (UTC)
Unfortunately weak and strong induction are logically equivalent, so it is invalid to say that: if it only works for weak induction, then it doesn't necessarily work for strong induction. So Resolution 2 is an invalid argument. --prime mover (talk) 13:43, 3 July 2014 (UTC)

It still doesn't quite hang together. The induction is claimed for $k \ge 1$. If $k \ge 1$ then $k + 1 \ge 2$ and so $T_c$ would be $\varnothing$ which is not an absurdity. --prime mover (talk) 14:11, 4 July 2014 (UTC)