= Solution
Reading the lower-left number in each source cell as player 1's <payoff>, against an opponent playing $A$ the player's own <payoffs> from $A,B$ are $y,0$; against $B$ they are $x,x/2$. For $x>0$, identify $A$ with aggressive defection and $B$ with yielding cooperation. The standard <payoffs> then have values
$$
T=x,\qquad R=x/2,\qquad S=0,\qquad P=y.
$$
The <prisoner's dilemma> ordering $T>R>P>S$ holds precisely for \b[$x>0$ and $0<y<x/2$] with this labeling. For example, $x=4,y=1$ gives $T=4,R=2,P=1,S=0$. Each player then has the <strictly dominant strategy> $A$, although $(B,B)$ gives both a larger <payoff> than $(A,A)$. Here $2R=T+S$ as well, so the optional weak aggregate-payoff condition is met with equality.
For the <Chicken payoff ordering>, the collision outcome must be worse than yielding to an aggressive opponent: $T>R>S>P$. This holds for \b[$x>0$ and $y<0$], for example $x=4,y=-1$. Against $B$, the <best response> is $A$ because $x>x/2$; against $A$, it is $B$ because $0>y$. Consequently $(A,B)$ and $(B,A)$ are the two <pure strategy> <Nash equilibria>, and mutual aggression is the worst outcome for both.
If action names may be exchanged, there is another <prisoner's dilemma> region: take $A$ as cooperation and $B$ as defection. Then $(T,R,P,S)=(0,y,x/2,x)$, so \b[$x<0$ and $x/2<y<0$] also work, for example $x=-4,y=-1$. This relabeling explains why the first displayed region is not a necessary condition without an action convention. Equality at any of these boundaries destroys the strict ordering.
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