Go backward to
Explicitly Defined Functions
Go up to
Top
Go forward to
Example
We prove
forall
x
:
x
' != 0.
Take arbitrary
x
.
By definition of 0 and ', it suffices to prove
x
union {
x
} !=
0
which is true because
x
in (
x
union {
x
}) but
x
!=
0
.
Author:
Wolfgang Schreiner
Last Modification: November 30, 1999