# Proving \$S=SRimplies S+phi\$

Let $$Lsubseteq Sigma^*$$ such that $${epsilon}notin L$$. Then for any $$Ssubseteq Sigma^*, S=SLimplies S=phi$$.

So we suppose $$S=SL$$ and $$Snephi$$. Then $$exists win S$$ such that $$0le|w|le |v|$$ for some $$vin S$$. Now $$|w|ne 0$$, since $$|w|=0implies{epsilon}=win S=SLimplies {epsilon}=xy$$ for some $$xin S$$ and $$yin Limplies x=y={epsilon}in L$$ which contradicts that $${epsilon}notin L$$.
Thus $$wne{epsilon}$$ and $${epsilon}notin S$$. So we have
begin{align} w in S &implies win SL\ &implies w=sl;text{for some}; sin S; text{and}; lin L\ &implies |w|=|s|+|l| end{align}
Here I’m stuck now. I think if I can show that $$|l|=0$$, then we can arrive at a contradiction.