Push down automata for $\{a^n b^n c^n | n \ge 0\}$

Question

I am learning about context free languages.

I understand how $\{a^n b^n c^n | n \ge 0\}$ can be shown to be not context free using the pumping lemma for CFL's.

Intuitively however it seems that a pushdown automata to recognize $\{a^n b^n c^n | n \ge 0\}$ can be constructed. This PDA would initially push two $a$'s into its stack whenever it sees an $a$ in the input. It would change state when it first encounters a $b$ and pop a single $a$. It would continue to pop $a$'s for every b in the input until it encounters a $c$. It would again change state and pop single $a$'s for every c encountered. If the stack is empty at the end of the input the language is recognized as $\{a^n b^n c^n | n \ge 0\}$.

There must be something I am overlooking whilst constructing the PDA as a language is context free if its has a PDA recognizing it. Please point out my mistake.

Answer 1

Your PDA would also accept strings of the form $a^nb^ic^j$ where $i+j=2n$ and $i,j>0$. For example the string "aaabbbbbc" is accepted. There is no way to tell that there are exactly $n$ $a$'s still remaining on the stack once you've finished reading the $b$'s.

Answer 2

The PDA would accept the strings which are not present in the languages . For example a string 'aabccc' can also be accepted by the PDA by popping out each a when it encounter b by changing the state anf when it encounters 'c' it will again pop out 'a' . Therefore PDA for a^nb^nc^n is not possible .

Answer 3

You are trying to say that for every a read from the tape you will push two a's on the stack then string aabccc will also be accepted by this PDA which is not part of given language