I-Does it true that having or not having elimination of quantifier property, say $EQ $, for structures depends on the language being used for describing the structure? If so, so one can "always" expand the language $L$ to $L'$, then the completion of a theory $T$, $T'$ will have $EQ$, isn't it? (I came across to this when I was reading $EQ$ for Discrete linear Orders and as far as I remember this method for proving $EQ$ has been used for some other theories too.)
II-May you please tell me an example to clarify this sentence" $EQ$ is equivalent to substructure completeness"?
Re: I, yes, quantifier elimination is extremely language dependent. Given any structure $\mathcal{M}$, there is a natural expansion $\mathcal{M}_{def}$ gotten by adding an $n$-ary relation symbol $R_\varphi$ for every $n$-ary formula $\varphi$ in the language of $\mathcal{M}$ which we interpret accordingly: $$R_\varphi^{\mathcal{M}_{def}}=\{(a_1,...,a_n)\in \mathcal{M}^n: \mathcal{M}\models\varphi(a_1,...,a_n)\}.$$ In one sense $\mathcal{M}_{def}$ doesn't add any essential strength to $\mathcal{M}$ - e.g. they have the same definable relations - but $Th(\mathcal{M}_{def})$is guaranteed to have QE whether or not $Th(\mathcal{M})$ has QE.
Re: II, the equivalence between substructure completeness and QE is treated here. Simply put, a first-order theory has QE iff it is substructure complete.