A's elapsed time is shorter than B's but B's is also shorter than A's.
How can you have two identical peices of string that are both the shorter of the two?
What you are missing is identifying the observer or the reference frame in your statements and questions. One thing relativity tells us is that identifying the reference frame is
crucial. It also tells us that measurements in different reference frames may come out differently; time looks different to observers that are moving relative to each other. You have not identified the frames properly.
In A's frame A's elapsed time is shorter than B's. In B's frame B's elapsed time is shorter than A's. There is no paradox. The measurements are made in different frames and need not agree. If you use the equations that relativity provides to transform the results from one frame to another you get consistent, agreeing answers.
But your second picture is drawn from the point of view of another observer, call her C, in a third frame. You didn't draw in observer C or identify her. Unintentionally introducing new and unidentified frames is a common cause of confusion in relativity. In C's reference frame, A is moving to the right and B is moving to the left, and each is moving with the same speed. From C's point of view, A's elapsed time is the same as B's elapsed time, and both are slower than C's elapsed time. In your second picture, t1 and t3 are both C's time, t2 is A's time, t4 is B's time, and t1 = t3 and t2 = t4 > t1.
If you want to just stick to A and B, you have to use either the first picture you posted (which is how the situation appears in B's frame) or use this one, a re-labeled mirror, that shows how the situation appears in A's frame:
Here we see the full symmetry between your first picture and this one; t5 = t2 and t6 = t1.
Always identify the frame or point of view when you make a statement. You can have two pieces of string that are identical when viewed in a reference frame in which they are not moving relative to each other, but piece A is shorter when viewed from a reference frame in which piece A is moving and piece B is not, and piece B is shorter when viewed from a frame in which piece B is moving and piece A is not.
{edited to correct a "B" to "A"}
This message has been edited by JonF, 12-07-2004 02:32 PM