Asklemmy

It comes down to how you define winning. Define L(X_i) as the 'loss' of warring party i at the end of the war - positive loss means that party i is worse off at the end of the war, while negative loss means party i is better off at the end of the war. If you are playing a board game, the rules might say someone always wins, and it is party i with the lowest L(X_i). But in a real life war, if party 1 started the war, their objective is probably that L(X_1) < 0 - i.e. they started the war to profit, not just to lose less than other parties. So in a real war, it is fair to say a party i loses if L(X_i) > 0, and wins if L(X_i) < 0. So to say no-one wins a war with parties P is to say \forall_{i \elem P} L(x_i) < 0.

Now in the case of wide scale nuclear war, parties likely launch all their nukes at each other within minutes so they launch before their capability to launch is destroyed. All major cities in all parties will likely be destroyed, and contaminated with nuclear fallout that may take years to decay to safe levels. Particulate thrown up by explosions would likely block out the sun and spoil all agriculture on earth for years (nuclear winter). Most people on earth would die. Government and civilisation would be unlikely to be able to continue under such circumstances - people might at least fall back to tribal organisation for a while.

So a widescale nuclear war would almost certainly lead everyone with a positive loss function - hence 'no winners'.