2. Every segment of ordinals (i.e., any set of ordinals arranged in natural order which contains all the predecessors of each of its elements) has an ordinal
number which is greater than any ordinal in the segment, and
3. The set
of all ordinals in natural order is well ordered.
Then by statements (3) and (1), has an ordinal . Since is in , it follows that by (2), which is a contradiction.
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of all ordinals in natural order is well ordered.
has an ordinal 
. Since 
is in 
, it follows that 
by (2), which is a contradiction.
