n

. The cases

n = 0, 1

P ([n - 1])

, consider one of its chains

A_{r} \subset A_{r + 1} \subset \dots \subset A_{n - 1 - r}, ∣ A_{j} ∣ = j .

It produces the two chains

A_{r} \subset \dots \subset A_{n - 1 - r} \subset A_{n - 1 - r} \cup {n}

and, when nonempty,

A_{r} \cup {n} \subset A_{r + 1} \cup {n} \subset \dots \subset A_{n - 2 - r} \cup {n} .

The first runs from rank

r

to rank

n - r

, and the second from rank

r + 1

to rank

n - 1 - r

; both endpoint ranks sum to

n

. They are disjoint and together contain the old chain both without and with

n

. Doing this for every old chain partitions

P ([n])

into symmetric chains.

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