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I have that language $S=\{a^n b^m c^m\mid n,m \geq 0\}$. How can I prove with the pumping lemma that it isn't regular? Can I use the concatenation closure and say that it's the language $L1 = \{a^n\mid n\geq0\}$ and $L_2 =\{b^m c^m\mid m \geq0\}$ prove that $L_2$ isnt regular so $L_1 L_2 = S$ is not regular too?

I have that language $S=\{a^n b^m c^m\mid n,m \geq 0\}$. How can I prove with the pumping lemma that it isn't regular? Can I use the concatenation closure and say that it's the language $L_1 = \{a^n\mid n\geq0\}$ and $L_2 =\{b^m c^m\mid m \geq0\}$ prove that $L_2$ isnt regular so $L_1 L_2 = S$ is not regular too?

I have that language S={w= a^n b^m c^m, n,m >=0} How can i prove with the pumping lemma that it isnt regular? Can i use the concatenation closure and say its the language L1 = {w=a^n, n>=0} and L2 ={w= b^m c^m, m >=0} prove that L2 isnt regular so L1 L2 = S is not regular too?

I have that language $S=\{a^n b^m c^m\mid n,m \geq 0\}$. How can I prove with the pumping lemma that it isn't regular? Can I use the concatenation closure and say that it's the language $L1 = \{a^n\mid n\geq0\}$ and $L_2 =\{b^m c^m\mid m \geq0\}$ prove that $L_2$ isnt regular so $L_1 L_2 = S$ is not regular too?