Proof of the One-Day Theorem In his paper John Conway has put forward some theorems describing the behaviour of the look-and-say sequence. The very first theorem, the One-Day Theorem, states that no one-day-old string contains substrings of the following types:
$•$ $yxzx$
$•$ $x^k$ $with$ $k ≥ 4$
$•$ $x^3y^3$
How to prove this theorem for the last two cases? The proof for the first case is that the sequence may be represented as $x^yx^z$, which simplifies into $x^{(y+z)}$, and as the superscript must always be the largest possible it is not allowed to divide it into more powers, i.e. $y$ and $z$. The other two are said to be special cases of the first one, but, frankly, I don't see how they are.
One-day old strings are the first iterations of the seeds in the look-and-say sequences, so if the seed is 2 then the one-day old string is 12. Look up this PDF for further clarifications (page 1-2).
 A: There are always exactly two ways to produce any sequence in look and say numbers. Either our sequence starts from a power or it starts from a value.
For example the sequence $x$ can only happen if day before we see: $a^x$ or $x^a$, where $a$ is any arbitrary value. In the first example $x$ starts from the power, and in the second one it starts from the value.
$$x^k, k \ge 4$$
First it should be noted that for any value $k \ge 4$ the sequence will always be a super sequence of $x^4$. Thus if we can show that $x^k$ can not exist when $k=4$ then that must be true for all values of $k>4$ as well.
Next lets look at if this value came from a power starting sequence, then the day before we would see $x^xx^x$. We can rewrite this sequence as $x^{2x}$ and that generates an $ax$, where $a=2x$ on the next day. Thus the sequence could not be generated starting from a power.
Lastly lets look at if this value came from a value starting sequence, then the day before we would see $x^ax^xb^x$, where $a$ and $b$ are arbitrary values. This can be rewritten as $x^{a+x}b^x$. This sequence would generate $cxxb$, where $c=a+x$. Thus it is impossible to generate $x^4$ from a value starting sequence, and by extension it is impossible to generate $x^k$, where $k\ge4$ in any circumstance.
What if $c=x$ and $b=x$ in our sequence $cxxb$?
Well that would be impossible because $c=x+a$ implies $a=0$ and we can not have $x^0$ as that would be read as nothing. We alternatively can show that if $b=x$, then on the day before we would have had $x^ax^xx^x$, which is rewritten as $x^ax^{2x}$, and that generates $axdx$, where $d=2x$.
$$x^3y^3$$
Our value starting sequence we get $x^ax^xy^yb^y$, where $a$ and $b$ are arbitrary values. This can be rewritten as $x^{a+x}y^yb^y$. This then generates the sequence $cxyyyb$, where $c=a+x$. Thus it can not be from a value starting sequence.
The day before, if this is from a power starting sequence we get $x^xy^xy^y$, which is rewritten as $x^xy^{x+y}$. This sequence would produce $xxcx$, where $c=x+y$. This shows that $x^3y^3$ is completely impossible.
A: As pointed out in the comments, the first should be no copies of $yxzx$ starting at an odd position. Crucially, $x,y,z$ do not have to be different here.
The other two are special cases. If we contain $a^4$ then either it starts at an odd position or an even position. If an odd position, we have $aaaa$ starting at an odd position so $x=y=z=a$. If an even position, then $a^4$ isn't at the start of the string, so we have $ba^4$ for some $b$. This contains $baaa$ starting at an odd position, here $x=z=a,y=b$.
If we contain $a^3b^3$ then either it starts at an odd position, giving $abbb$ starting odd (here $x=z=b,y=a$), or an even position, and looking at the previous position gives $caaa$ starting at odd for some $c$.
