Show that language is not a regular language. L = 0m 1m mid m geq 1

--- If A is a regular language, then there exists a pumping length p such that any string s with s ≥ p can be divided into three parts s = xyz satisfying: 1. xy^i z ∈ A for every i ≥ 0 2. y 0 (y cannot be empty) 3. xy ≤ p (xy together must not exceed pumping length) --- Step 1: Assume L is regula...