Using the property that the intersection of a CFL and a regular language is still a CFL, and the fact that the language

{a^n b^n c^m | 0 m n} is not context free, prove that the following language A is not context free.

A = {w | w , and in w, the number of as is equal to the number of bs and is larger than or equal to the number of cs},

(i.e., aabcb and cba are in A. baac or accb are not in A.)