Solutions to the Set Theory Exercises
2. Defining union
Definition: y = Uz <---> Ax(x in y <---> Ew (w in z &
x in w)).
Existence proof:
- Az Ey Ax (x in y & Ew (w in z & x in w) ---> x in y) [union
axiom, AE]
Uniqueness proof:
- | (x in y <---> Ew (w in z & x in w))
- | (x in y' <---> Ew (w in z & x in w))
- | x in y <---> x in y'
- | Ax (x in y <---> x in y')
- | Ax (x in y <---> x in y') ---> y = y' [axiom of extensionality]
- | y = y'
- (x in y <---> Ew (w in z & x in w)) ---> y = y'
Equational Definitions
Let t(x) be an introduced term.
Generic proof of existence for a definition in terms of a term
- y = t(x)
- Ez y = z [EI]
Generic proof of uniqueness for such a definition
- | y = t(x) & y' = t(x)
- | y = y' [=]
- y = t(x) & y' = t(x) ---> y = y'
From now on, you don't have to justify definitions in terms of identity
with terms.
Incidentally, a definition in terms of a term may be equivalently rephrased
as an equation:
is logically equivalent to
Proof:
- | f(x) = t(x)
- | | f(x) = y
- | | y = t(x) [=]
- | f(x) = y ---> y = t(x)
- | | y = t(x)
- | | f(x) = y
- | y = t(x) ---> f(x) = y
- | y = t(x) <---> f(x) = y
- f(x) = t(x) ---> (y = t(x) <---> f(x) = y)
- | y = t(x) <---> f(x) = y
- | Ay(y = t(x) <---> f(x) = y) [AI]
- | t(x) = t(x) <---> f(x) = t(x)) [AE]
- | t(x) = t(x)
- | f(x) = t(x)
- (y = t(x) <---> f(x) = y) ---> f(x) = t(x)
- (y = t(x) <---> f(x) = y) <---> f(x) = t(x)
1. Defining Intersection
Intersect(z) = {x in Uz: Aw(w in z ---> x in w)}.
3. Defining x U y
x U y = U{x, y}
4. Replacement: as usual.
5. Show that
-Es s is a function & Ax, y, (x, y> in s <---> u = x U
{x}
Proof
- | Es s is a function & Ax, y, (<x, y> in s <---> y
= x U {x}) [hyp]
- | | s is a function & Ax, y, (<x, y> in s <---> y =
x U {x}) [hyp]
- | | x in dom(s) <---> Ey <x, y> in s [defn dom]
- | | x in dom(s) <---> Ey y = x U {x} [logic]
- | | x in dom(s) [logic]
- | | Ax x in dom(s) [logic]
- | | Ey Ax x in y [EI]
- | Ey Ax x in y [EE]
- | _|_ [Russell's paradox using separation]