in Theory of Computation
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Prove that $(L_1L_2)^R=L_2^RL_1^R$
for all languages $L_1$ and $L_2$.
in Theory of Computation
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$(L_1L_2)^R = \{ x^R | x \in L_1L_2 \}$

Since $x \in L_1L_2,$ hence, $x = yz; \,\, y \in L_1, z \in L_2$

$(L_1L_2)^R = \{ (yz)^R | y \in L_1, z \in L_2 \}$

$(L_1L_2)^R = \{ z^Ry^R | y \in L_1, z \in L_2 \}$  (by the result we proved Here )

$(L_1L_2)^R = \{ z^R |  z \in L_2 \}.$ $\{ y^R | y \in L_1 \}$

$(L_1L_2)^R = L_2^R.L_1^R.$

Hence Proved.

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