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Free shipping on millions of items Get the best of Shopping and Entertainment with Prime Enjoy low prices and great deals on the largest selection of everyday essentials and other products, including fashion, home, beauty, electronics, Alexa Devices, sporting goods, toys, automotive, pets, baby, books, video games, musical instruments, office supplies, and more( b ( a # c $ ( &Since apbapbapb=s=xyz=ajakambapbapb=ajkmbapbapb, we must have that jkm=p The first condition implies that xy2z∈ A1, but xy2z = ajakakambapbapb = apkbapbapb since j km = p Hence, xy2z ∈ A1 because k ≥ 1, and we get a contradiction Therefore, A1 is a nonregular language (b) ={w∈ {a,b}∗ w=wR} Answer Suppose that is

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