# Propositions logic and problem solving

How can a question of this nature be approached:

Two avid game players Alice and Bob play three diﬀerent games. They are very competitive and so would do anything within the rules of the game to win. Each player chooses an element $$x$$ or $$y$$ from a set (domain of discourse) $$D$$, according to certain rules. The outcome of the game is decided by a propositional function $$P(x,y)$$. If $$P(x,y)$$ holds, then Bob wins. If $$P(x,y)$$ fails then Alice wins. Suppose that as long as the players obey the rules, Bob always wins.

Describe a proposition that is true corresponding to Bob winning each game.
(a) Alice plays $$x$$ ﬁrst, then Bob is allowed to play $$y$$.
(b) Alice plays both $$x$$ and $$y$$.
(c) Bob plays $$x$$ then Alice plays $$y$$.

And what are the steps to coming up with a solution?

• the games seems to favour Bob so if Bob chooses anything he wins $\forall x B(x)$ (B(x) nmeaning Bob chooses x) Oct 28, 2014 at 10:09

Consider (a). From the fact that Alice plays $x$, we deduce that $x$ must be better than all other options from $D$. So since Bob can win if Alice chooses $x$, the only reason to choose $x$ over some $x'$ is that for $x' \ne x$, Bob must win. That is:
$$\forall x': \forall y': x' \ne x \to P(x',y')$$
On the other hand, since Bob plays $y$, we know that there is no other choice for $y$, which is expressed as:
$$\forall y': P(x,y') \leftrightarrow y = y'$$
These two together fully determine $P$; I leave it to you to describe it explicitly.
For (b), we can deduce similarly that $x$ and $y$ are equally valid options for Alice, while all others are strictly worse (i.e. are guaranteed to lose). We are at liberty to decide whether Bob must or only can win. The reasoning for (c) is entirely analogous.