Right now my tires in the front are new. If I measure to the center line of the wheel I have a diameter of 27" diameter. With that diameter my wheel has a circumference of about 84 inches. If I travel ten revolutions of the tire I cover 840".
Now lets "wear" an inch off the tread. This changes the diameter of the tire to 25". The wheel now travels 25 x 3.14152 inches or 78" per revolution. Ten rotations now covers 780".
The connection between the front axle and the back axle is "locked" when in 4wd. That being said, the drive shaft into the front axle must match the rotation of the rear axle. If both connections are rotating at the same speed then the engage/disengage is a easy task. Now imagine the front axle with the 1" or wear on the tires. The front wheels are now rotating faster to cover the original design (new tires) of 840". The rear tire size has not changed any appreciable amount (fronts wear out way faster than rears). This causes a mismatch of the rotation speeds where the front axle meshes with the rear axle thus a bind is felt when shifting.
You can travel in 4wd all the time but as the front tires wear the ratio becomes more of a mismatch. when locked in 4wd this "mismatch" of turning must be accounted for. Since gears can't "slip", the only place that the "slip" can occur is at the contact with the ground. With the rears being heavier than the fronts, the fronts are forced to make up the mismatch by sliding on the ground slightly, this sliding accelerates the wear of the front tires thus accelerating the change in the diameter.
I am sure we are both on the same page that this all comes down to is get your tractor out of 4WD unless you need 4WD and keep your tires inflated fully, especially as they get shorter. By all means get it out of 4WD when you have a heavy load in the bucket on a solid surface where the front tires cannot "slip" as would happen on gravel.
Hope this makes some sense John. I suspect we are on the same page just a different language.
Gregg