Type Assertion in TypeScript without assigning a new variable

I have a function that returns values with one of two interfaces, like so:

interface K {
    A:number;
    NUM:number;
}

interface L {
    A:string;
    STR:string;
}

function FUN():K|L { }

var x:K|L = FUN();

but I can't access the non-shared property using a type-guard like this one:

if(typeof x.A === 'string'){
    console.log(x.STR) // Property 'STR' does not exist on type 'K | L'
}

even though the if clause does distinguish between the two types. I know I can use as operator with a variable assignment to indicate the type to the compiler like so:

if(typeof x.A === 'string'){
    var y = x as L;
    console.log(y.STR) // OK 
}

but that seems a bit kludgy as this add a unnecessary var statement and assignment in the resulting js code:

function FUN() { }
var x = FUN();
if (typeof x.A === 'string') {
    var y = x; // unnecessary...
    console.log(y.STR); 
}

Is there any way to indicate the type in this situation without the extra var y=x; in the JS code?