 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 valuesThe prisoner's dilemma ordering $T>R>P>S$ holds precisely for $x>0$ 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 $x>0$ 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 $x<0$ 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.