fix(types): allow $cond with $in

Fix #12028

Author: vkarpov15

test/types/expressions.test.ts

@@ -174,7 +174,15 @@ const addWithNull: Expression.Add = {
     '$price',
     { $ifNull: ['$tax', 0] }
   ]
-}
+};
+
+const condWithIn: Expression.Cond = {
+  $cond: {
+    if: { $in: [] },
+    then: '$foo',
+    else: '$bar'
+  }
+};
 
 const toLong: Expression = { $toLong: '$qty' };
 

types/expressions.d.ts

@@ -1043,7 +1043,7 @@ declare module 'mongoose' {
        *
        * @see https://docs.mongodb.com/manual/reference/operator/aggregation/cond/#mongodb-expression-exp.-cond
        */
-      $cond: { if: BooleanExpression | ConditionalExpressionOperator, then: AnyExpression, else: AnyExpression } | [BooleanExpression, AnyExpression, AnyExpression];
+      $cond: { if: Expression, then: AnyExpression, else: AnyExpression } | [BooleanExpression, AnyExpression, AnyExpression];
     }
 
     export interface IfNull {