In the following tutorial question and solution, am I correct in thinking that the logic is flawed for the following reason? It was not explicitly stated that $u$ is not quasiconcave or quasiconvex, thus, it is too strong an assumption to assume that $u(\gamma s_1 + (1-\gamma)s_2)\neq u(s_1, \sigma_{-i}^*)\quad\text{or}\quad u(s_2, \sigma_{-i}^*)$

Tutorial question

To give a brief introduction to the concepts, $I$ players are participating in a "game" where they each choose a strategy $s_i$ in response to the other players' strategies $\sigma_{-i}$. $\quad\sigma_i$ represents a mixed strategy, and $u$ is the payoff to each player.

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    $\begingroup$ $A \to \lnot (B \land \lnot A)$ is Always TRUE. Thus, it is implied by whatever proposition and so : YES, it is implied by $A \to B$. $\endgroup$ Apr 9, 2019 at 15:50
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    $\begingroup$ Can the poster change the title of this question? It is grossly misleading. $\endgroup$
    – max_zorn
    Apr 9, 2019 at 21:18

1 Answer 1


The logic in the solution is correct. $$u(\gamma s_1 + (1-\gamma)s_2) = u(s_1, \sigma_{-i}^*) = u(s_2, \sigma_{-i}^*)$$ does contradict "$u$ is strictly quasiconcave".

To see this, let $A$ be the statement "$u$ is strictly quasiconcave", and $B$ be the statement "$u$ is quasiconcave". Then $A\implies B$, but if $A$ is true, then $(B\wedge \neg A) =$ $$u(\gamma s_1 + (1-\gamma)s_2) = u(s_1, \sigma_{-i}^*) = u(s_2, \sigma_{-i}^*)$$ is not true.


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