A Markov algorithm that does unary multiplication

I am learning Markov algorithms and came across the paper Markov Algorithm by CHEN Yuanmi. I am trying to understand the following example (well, any of the examples in the paper).

Example 3: Multiplication. MA3 = $\{\Sigma, P, P_0\}$

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$\Sigma = \{a, A, B, C, \#\}$

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Productions in $P$:

$p_1: Ba \to aB$

$p_2: Aa \to aBA$

$p_3: a \to \epsilon $

$p_4: a\#\to \#A $

$p_5: \#a \to \#$

$p_6: \# \to \epsilon$

$p_7: B \to a$

$p_8: \epsilon \to \cdot \epsilon$

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$ P_0 = \{p_8\}$

Remark: Given the input in form of "$\underbrace{aa...a}_x\#\underbrace{aa...a}_{y}$", the algorithm will halt with the output in the form of "$\underbrace{aa...a}_{x \times y}$".

My question is about the above example. I understand the rules should be applied in linear order. So given an input of "$aaa\#aaa$", the first rule to match would be $p_3$, which says to replace a with the empty character. So the string will be consumed down to a single #, which is then also replaced with an empty character ($p_6$), then the machine halts with no output($p_8$). How does this achieve multiplication? What am I not understanding?

You have demonstrated that example on multiplication cannot be correct.

Well, there is typo in that example. The substitution rule $p_3$ should have been
$$A\to\epsilon$$

StackExchange Computer Science Q#156433, answer score: 2