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0 := 0;x' :=xunion {x}.

*Proof* of first Peano law:

We prove **forall** `x`: `x`' != 0. Take arbitrary `x`. By
definition of 0 and ', we have to prove

xunion {x} !=0

which is true because `x` in (`x` union {`x`}) but
`x` not in **0**.

*Proof* of second Peano law: see lecture notes.

Author: Wolfgang Schreiner

Last Modification: November 16, 1999