L={a^n,b^n,c^m / n>m}
we have to compare n and m everywhere I know till we reach b there will be nothing in the stack BUT generally we can make this language if possible
suppose when we are putting one a into the stack suppose we put Two a's on behalf of one a then the no'of a's 2*n into the stack and the we pop one a for one b after completing allb's still there will be n number of a's will be remaining into the stack then we can compare number of a's and number of c's if and one a or more than one a is remaining in the stack then we can accept the string otherwise we can reject the string
I'm asking this question because in one of the video for the question we were taking two a's on behalf of one .......