There is nothing in Banesh Hoffmann's text above that justifies the words "without contradiction". "Without contradiction" could be justified if Einsteinians had devised and discussed at least one scenario where EITHER observer's clock is checked against two of the other observer's clocks (at some distance from one another). Instead, discussions in Divine Albert's world are restricted to the misleading twin paradox scenario where the travelling twin is implicitly deprived of the setup "two clocks at some distance from one another against which the sedentary twin's clock could be checked" and so there is no way to demonstrate the slowness of the sedentary twin's clock. Only the slowness of the travelling twin's clock can be demonstrated, Divine Einstein, yes we all believe in relativity, relativity, relativity.
The following scenario allows EITHER observer's clock to be checked against two of the other observer's clocks. Two long inertial systems each carrying synchronous clocks pass one another:
..........Inertial system A moving to the right.......... ..........Inertial system B moving to the left..........
The systems are so designed that, the moment they stop moving relative to one another, all clocks on both systems stop ticking. In this final static configuration clock A2 faces clock B1 and clock A1 faces clock B2:
Before reaching clock A2, clock B1 passed clock A1 and the difference in their readings, (A1then - B1then), was then registered. *Now*, in the final static configuration, clock B1 faces clock A2 and the difference in their readings is (A2now - B1now). Clearly clock B1 has been checked against two of Inertial system A's clocks so, according to special relativity, the following inequality holds:
(A2now - B1now) > (A1then - B1then) /1/
Before reaching clock B2, clock A1 passed clock B1 and the difference in their readings, (B1then - A1then), was then registered. *Now*, in the final static configuration, clock A1 faces clock B2 and the difference in their readings is (B2now - A1now). Clearly clock A1 has been checked against two of Inertial system B's clocks so, according to special relativity, the following inequality holds:
(B2now - A1now) > (B1then - A1then)
This inequality easily becomes:
(A1then - B1then) > (A1now - B2now)
Since clocks on Inertial system A were synchronous and stopped ticking simultaneously, A1now = A2now. For the same reason B2now = B1now. So the last inequality becomes:
(A1then - B1then) > (A2now - B1now) /2/
Inequalities /1/ and /2/ are contradictory and both are consequences of Einstein's 1905 light postulate. Reductio ad absurdum par excellence. The light postulate is false.