Is the set of pairs of TMs at least one of which accepts the empty word semi-deciable? L = { &lt; M1,M2 > | M1,M2  are TM's and Ɛ ∈ L(M1) ∪ L(M2) }

Where Ɛ = Epsilon 

I know that this language is undecidable, but why it is semidecidable too.

What i have tried is =>

Using Rice's Theorem, part 2 

T(no) = (0,1)^+ and T(yes) = (0,1)*

Which gives that T(yes) is not a subset of T(no), and this says that it is semidecidable. 

Is my understanding right ?
Rice's second theorem can be used to show for some sets that they are not semi-decidable, but not the opposite. 

In order to show that this set is semi-deciable, just follow the definition and construct a semi-decider.


   Idea: Simulate both $M_1$ and $M_2$ on $\varepsilon$ simultaneaously using dovetailing. If either one halts and accepts, stop simulating and accept.