$$ G \subset J \quad \text{and} \quad \langle \{\mathrm{LM}_{\geq}(g) \;|\; g \in G \} \rangle = \langle \{\mathrm{LM}_{\geq}(f) \;|\; f \in J\} \rangle. $$

Such a definition implies that $G$ is another generating set of $J$, i.e. $\langle G \rangle = J$. Hence if we have a Groebner basis $G$ of $J$ and a subset $G' \subset J$ which we would like to test for being a Groebner basis of $J$, we need to verify that

$$ G' \subset \langle G \rangle \quad \text{and} \quad \langle \{\mathrm{LM}_{\geq}(g') \;|\; g' \in G' \} \rangle = \langle \{\mathrm{LM}_{\geq}(g) \;|\; g \in G \} \rangle. $$

In order to test whether $G' \subset \langle G \rangle$ we can use another property of a Groebner basis which states that

$$ f \in \langle G \rangle \iff \overline{f}^{(G, \geq)} = 0, $$

where $\overline{f}^{(G, \geq)}$ is the remainder of the division of $f$ by $G$ w.r.t. $\geq$. As for the equality of monomial ideals (i.e. ideals generated by monomials), if we have two sets of monomials $M_1$ and $M_2$, then

$$ \langle M_1 \rangle \subset \langle M_2 \rangle \iff \forall \mathbf{x}^{\boldsymbol{\alpha}} \in M_1: \;\exists \mathbf{x}^{\boldsymbol{\beta}} \in M_2: \;\mathbf{x}^{\boldsymbol{\beta}} \;|\; \mathbf{x}^{\boldsymbol{\alpha}}, $$

where the notation $\mathbf{x}^{\boldsymbol{\beta}} \;|\; \mathbf{x}^{\boldsymbol{\alpha}}$ means that $\mathbf{x}^{\boldsymbol{\beta}}$ divides $\mathbf{x}^{\boldsymbol{\alpha}}$. The equality $\langle M_1 \rangle = \langle M_2 \rangle$ can be then checked by verifying that the two inclusions hold true.