Let $1 \leq m \leq n, \sigma \in S_n, \pi \in S_m$. The permutation $\sigma$ avoids $\pi$ if no subset $\{j_1 < \cdots < j_m\} \subseteq \{1,\cdots,n\}$ exists, so that for all $1 \leq i < l \leq m$ applies $\sigma(j_i) < \sigma(j_l) \Leftrightarrow \pi(i) < \pi(l)$
Prove that if a permutation $\sigma \in S_n$ avoids the permutation 123 it avoids the permutation 3124 as well (one-line-notation used for permutations).
Hi!
I first reflected on what the definition of the avoiding implies for $\sigma$ (apart from $|3124| = m = 4 \Rightarrow n \geq 4$). The only fact I discovered is that according to the definition $\sigma$ contains one ascension at most and this should "cover" the last part $124$ of $3124$.
Could you please help me to go on?
Thanks in advance!