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If L1 and L2 are Turing-Recognizable then L1 ∪ L2 will be decidable???

can i say L1 and L2 is re and re is closed under union operation so it is decidable??

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$L_{1}=RE$

If $L_{2}=RE$and complement of $L_{1}$ then only it will be decidable
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RE $\cup$ RE = RE
If $\bar{L1}$ and $\bar{L2}$ are also RE then only their union is Decidable i.e Recursive