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$(uv)^R = v^Ru^R,$ for all strings $u,v.$

If $u = \epsilon,$ then $u^R = \epsilon$ and hence $(uv)^R = v^R = v^Ru^R.$

If $v = \epsilon,$ then $v^R = \epsilon$ and hence $(uv)^R = u^R = v^Ru^R.$

Now, suppose $u = u_1u_2 \dots u_m$ and $v = v_1v_2 \dots v_n$, with $m,n \geq 1.$

Then, $(uv)^R = (u_1u_2 \dots u_mv_1v_2 \dots v_n)^R = v_n \dots v_2v_1x_m \dots x_2x_1 = v^Ru^R.$

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Naveen Kumar 3 asked Mar 19, 2019
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Prove or disprove the following claims.(a) $(L_1 ∪ L_2)^R = L_1^R ∪ L_2^R$ for all languages $L_1$ and $L_2$.(b) $(L^R)^* = (L^*)^R$ for all languages $L$.