# If $A\implies B$, then $A\implies \neg(B\wedge\neg A)$

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 $$u(\gamma s_1 + (1-\gamma)s_2)\neq u(s_1, \sigma_{-i}^*)\quad\text{or}\quad u(s_2, \sigma_{-i}^*)$$

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.

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

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.