Solution af904b52-a1e4-46f9-9f56-63055a74601e

Points: 3 pts

Lemma 1: If there were only one tiger and one sheep, the tiger would eat it.

Let $n$ be the number of tigers. Using $1=\text{true}$ and $-1=\text{false}$, define the truth value of
``The sheep will not be eaten'' by
\[
f(n)=
\begin{cases}
1, & \text{if the sheep will not be eaten},\\
-1, & \text{if the sheep will be eaten}.
\end{cases}
\]
At $n=1$, if the tiger eats the sheep, there are no more tigers, so the new sheep will not
be eaten. It is therefore rational for the tiger to eat the sheep, so
\[
f(1)=-1.
\]

Lemma 2: $f(n)=-f(n-1)$.

For example, if there were two tigers $(n=2)$, neither tiger would eat the sheep. If one
did, there would be one tiger remaining who could eat the new sheep safely. We see that
\[
f(2)=-f(1)=1.
\]
If there were three tigers $(n=3)$, a tiger could safely eat the sheep. If one did, the other
two tigers would remain passive, because any tiger that ate the new sheep would then be
eaten. Thus
\[
f(3)=-f(2)=-1.
\]
In general, a tiger can safely eat the sheep precisely when it knows that, once it becomes
the sheep, it will not be eaten. If the new sheep would be eaten, all tigers remain passive.
The two truth values are therefore opposites. Boolean negation swaps $1$ and $-1$, giving
\[
f(n)=-f(n-1)\qquad(n\geq 2).
\]

Lemma 3: $f(n)=(-1)^n$.

We solve the recurrence $f(n)=-f(n-1)$, starting from $f(1)=-1$:
\[
f(n)=(-1)^{n-1}f(1)=(-1)^n.
\]
We can then substitute $n=100$:
\[
f(100)=(-1)^{100}=1.
\]

Final answer:
\text{No}