Adding new members of minimal military value in order to consciously antagonize the one power that is supposed to be threatening Europe is a bizarre tactic.
It is important to note that in the present experiment, the recognition of words with long first syllables was significantly facilitated by falling tone despite the fact that the spoken stimuli were constructed on the basis of the short member of the minimal quantity pair.
In what follows, we show that (S^{mathrm{sl}} ({mathcal{E}_{0} ^{ast} })) has at least a minimal member, and (bar{mathcal{E}}) is a minimal member of (S^{mathrm{sl}} ({mathcal{E}_{0} ^{ast} })) if and only if (bar{mathcal{E}} in S^{mathrm{l}} cap S^{mathrm{sl}} ({mathcal{E}_{0} ^{ast} })).
In [8], Huang and Jia applied the basic Lemma 2.3 and the famous Zorn lemma to show the existence of minimal members in the course of the proof of their other main theorem (Theorem 3.1) (see line 33 of p.48).
(mathcal{E}_{0} ) is a minimal member of S; (mathcal{E}_{0} ) is a linear expectation.
(mathcal{E} ) is a minimal member of (S^{mathrm{sl}}); (mathcal{E} ) is a linear expectation.
Then E is a minimal member of (S^{mathrm{cv}}) and a maximal member of (S^{mathrm{conca}}).
Then the following statements are equivalent: (i) (mathcal{E} ) is a minimal member of (S^{mathrm{sl}}); (ii) (mathcal{E} ) is a linear expectation. .
We show that (mathcal{E}) is a minimal member of S if and only if (mathcal{E}) is a linear expectation with the same constraints as above, respectively.
Then the following two statements are equivalent: (i) (mathcal{E}_{0} ) is a minimal member of S; (ii) (mathcal{E}_{0} ) is a linear expectation. .
