The following presents properties of the TSNE in relation with domination and strict NE:

**Theorem 2::** *The following hold*:

We wish to discuss direct implications of and issues about this theorem before its proof.

First, it should be emphasized that when evaluating the performance of our notion against strict domination, Theorem 2–1 ensures that we do not encounter the type of problems often cited in the discussion of “imperfections of perfection” (see myerson (1978)).
To see this, consider the following:

**Example 4::** *First, consider the strict dominance truncation of the following game*:

*The NE are $s=(I,I)$ and ${s}^{{}^{\prime}}=(II,II)$, but only $s$ is perfect. But considering strictly dominated strategies as well results in $II$ not being weakly dominated. NE and perfect equilibria coincide and are equal to $s=(I,I)$ and ${s}^{{}^{\prime}}=(II,II)$. But in both cases $\mathbb{T}(s)=(1,1)$ and $\mathbb{T}({s}^{{}^{\prime}})=(2,2)$, so the TSNE equals $\{s\}$*.

The second point concerns our finding that the TSNE exhibits stronger refinement powers than the other refinements of NE whenever the game at hand possesses a strict NE. Indeed, Theorem 2–2 immediately implies (1) the TSNE contains neither weakly dominated NE nor mixed NE and is a strict subset of the lower hemi continuous selection of NE;
and (2) the TSNE is a strict subset of the set of perfect equilibrium, proper equilibrium, and PE. While the battle of the sexes shows that the containment relation of the TSNE with perfection and properness is strict, the following example performs the same task with persistence:

**Example 5::** *This game is one that has a strict NE but no redundant and weakly dominated actions, and the TSNE is a strict subset of the set of PE*.

*Here*, ${s}^{1}=(I,I)$ *is a strict NE which, therefore, is not empty. Thus, the TSNE equals the set of strict NE, hence, does not contain the mixed strategy NE*, ${s}^{2}=(1/2II+1/2III,1/2II+1/2III)$*. But* ${s}^{2}$ *is a PE*:$R={\times}_{i=1,2}{R}_{i}$ *defined by* ${R}_{i}=\{(0,x,1-x)\text{\hspace{0.17em}}:\text{\hspace{0.17em}}x\in [0,1]\}$, $i=1,2$*, is an essential Nash retract because (1) for* $\mathrm{\epsilon}>0$ *sufficiently small the best-responses of agents against* $(\mathrm{\epsilon},x,1-x-\mathrm{\epsilon})$ *do not contain* $I$*; (2) it is minimal*.

On the other hand, when there is no strict NE, example 1, a game with no redundant and weakly dominated actions, shows both a pure strategy and a mixed strategy NE may be in the TSNE and this notion may not be lower hemi continuous.
Moreover, the TSNE may contain weakly dominated NE when the game has no strict NE. This is due to the following:

**Example 6::** *This game has no strict NE and no redundant actions, but two NE*: $s=(II,II,II)$ *and* ${s}^{{}^{\prime}}=(1/2I+1/2II,1/2I+1/2II,I)$.

$s$*, involving a weakly dominated action by player 3, is in the TSNE*: $\mathbb{T}(s)=(1,1,2)$ *and* $\mathbb{T}({s}^{{}^{\prime}})=(2,2,1)$.

The TSNE may not eliminate weak domination because when there is no strict NE it may not be able discriminate between weak domination and randomization: in example 6, $\mathbb{T}{}_{3}(s)=2$ because ${s}_{3}=II$ is a weakly dominated action for player 3; ${\mathbb{T}}_{i}({s}^{{}^{\prime}})=2$ because ${s}_{i}^{{}^{\prime}}=1/2I+1/2II$ for $i=1,2$ is totally mixed. But weak domination is not permitted with persistence (Kalai and Samet 1984, Theorem 4, p.139) due to the minimality requirement of essential Nash retracts. Therefore, the global evaluation measure embedded in persistence results in the elimination of weak domination, while the local means of evaluation with the TSNE does not suffice towards this regard.

**Proof of Theorem 2::** *The first item of the above theorem stated formally is: For any $\mathrm{\Gamma}\in \mathcal{G}$, $\mathcal{T}(\mathcal{N}(\mathrm{\Gamma}))\stackrel{\mathcal{D}}{=}\mathcal{T}(\mathcal{N}(\mathcal{D}(\mathrm{\Gamma})))$. Because that for any $\mathrm{\Gamma}$ we have that $\mathcal{N}(\mathrm{\Gamma})\stackrel{\mathcal{D}}{=}\mathcal{N}(\mathcal{D}(\mathrm{\Gamma}))$, we prove that for any $s\in \mathcal{N}(\mathcal{D}(\mathrm{\Gamma}))$ and ${s}^{{}^{\prime}}\in \mathcal{N}(\mathrm{\Gamma})$ with ${s}^{{}^{\prime}}\stackrel{\mathcal{D}}{=}s$ it must be that ${\mathbb{T}}_{i}^{\mathrm{\Gamma}}({s}^{{}^{\prime}})={\mathbb{T}}_{i}^{\mathcal{D}(\mathrm{\Gamma})}(s)$ for all $i\in N$. Then, the definition of the TSNE implies that $s\in \mathcal{T}(\mathcal{N}(\mathcal{D}(\mathrm{\Gamma})))$ if and only if ${s}^{{}^{\prime}}\in \mathcal{T}(\mathcal{N}(\mathrm{\Gamma}))$ where ${s}^{{}^{\prime}}\stackrel{\mathcal{D}}{=}s$. The desired conclusion follows from $\mathcal{P}{\mathcal{B}}_{i}^{\mathrm{\Gamma}}({s}_{-i}^{{}^{\prime}})\stackrel{\mathcal{D}}{=}\mathcal{P}{\mathcal{B}}_{i}^{\mathcal{D}(\mathrm{\Gamma})}({s}_{-i})$ because $\mathcal{D}(\mathrm{\Gamma})$ is a strict dominance truncation of $\mathrm{\Gamma}$, and on account of being a NE ${s}_{-i}$ and ${s}_{-i}^{{}^{\prime}}$ do not assign strictly positive probabilities to strictly dominated strategies, and player $i$ cannot assign strictly positive probabilities to strictly dominated actions in his best response.*

In order to prove item 2 we show: Let $\mathrm{\Gamma}\in \mathcal{G}$ be such that ${\mathcal{N}}_{s}(\mathrm{\Gamma})\ne \mathrm{\varnothing}$; then, $\mathcal{T}(\mathcal{N}(\mathrm{\Gamma}))={\mathcal{N}}_{s}(\mathrm{\Gamma})$. This follows from (1) the observation that for any strict NE, ${s}^{\ast}$, it must be that ${\mathbb{T}}_{i}({s}^{\ast})=1$ for all $i\in N$; and (2) for any NE that is not strict, ${s}^{{}^{\prime}}$, there exists $j\in N$ such that ${\mathbb{T}}_{j}({s}^{{}^{\prime}})>1$.

This finishes the proof of Theorem 2. ■

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