Snookered After A 'Free Ball'

[DawRef]

The rules state that you cannot snooker behind a 'free ball' unless only Pink and Black remain. The rules also state that:
If the cue-ball is so obstructed from hitting a ball on by more than one ball not on:
(i) the ball nearest to the cue-ball is considered the effective snookering ball, and
(ii) should more than one obstructing ball be equidistant from the cue-ball, all such balls will be considered to be effective snookering balls.

So, in (ii) above, what if one of those balls were a free ball, would you call a foul?

My answer is "Yes", because if you removed the other ball not on, the object ball would still be snookered by the free ball (i.e. you would not be able to strike both extreme edges of the ball on).
Does anyone else agree (or disagree)?

[Sarmu]

I would agree with a foul since in the case of (ii), all equidistant balls have to be considered, and if one of those were a free ball, then the cannot snookered behind a free ball rule should come into play

[gingerale]

If the described situation is meant unless only Pink and Black remain I'd agree in calling a foul.
Except
16(c) When Red is the ball on, if the cue-ball is obstructed from hitting different Reds by different balls not on, there is no effective snookering ball.
of course.

[The Statman]

The rules state that you cannot snooker behind a 'free ball' unless only Pink and Black remain. The rules also state that:
If the cue-ball is so obstructed from hitting a ball on by more than one ball not on:
(i) the ball nearest to the cue-ball is considered the effective snookering ball, and
(ii) should more than one obstructing ball be equidistant from the cue-ball, all such balls will be considered to be effective snookering balls.

So, in (ii) above, what if one of those balls were a free ball, would you call a foul?

My answer is "Yes", because if you removed the other ball not on, the object ball would still be snookered by the free ball (i.e. you would not be able to strike both extreme edges of the ball on).
Does anyone else agree (or disagree)?

Absolutely agree. Both balls are 'snookering balls' and if one of them is the free ball, that is unequivocally against the rules.