Math 99 (Notes) encoding

To Prove:		Do:
P\implies Q	"Assume P is true," prove Q is true, or "Assume Q is false," prove P is false, or "Assume P is true and Q is false", produce a contradiction.
P\iff Q	Prove (P\implies Q)\And(Q\implies P), or prove (P\implies Q)\And(\Not P\implies\Not Q), or prove (\Not Q\implies\Not P)\And(Q\implies P), or prove (\Not Q\implies\Not P)\And(\Not P\implies\Not Q)
(\forall x)(P(x))	"Let x be an arbitrary \ldots" Prove P(x).
(\exists x)(P(x))	"Take x = \ldots" Prove P(x) for this x.
A \subseteq B	Prove (\forall x\in A)(x\in B) i.e., if x\in A then x\in B.
A = B	Prove (A \subseteq B) \And (B \subseteq A).
A = \emptyset	Prove (\forall x)(x\notin A) (frequently best to use proof by contradiction).
x\in A \cup B	Prove (x\in A) \Or (x\in B).
x\in A \cap B	Prove (x\in A) \And (x\in B).
x\in A - B	Prove (x\in A) \And (x\notin B).
\Not(P(x) \implies Q(x))	Prove (\exists x)(P(x)\And\Not Q(x)).
\Not (P(x) \iff Q(x))	Prove (\exists x)(P(x)\And\Not Q(x))\Or(\exists x)(Q(x)\And\Not P(x)).
\Not(\exists x)(P(x))	Prove (\forall x)(\Not P(x)).
\Not(\forall x)(P(x))	Prove (\exists x)(\Not P(x)).
A\not\subseteq B	Prove (\exists x)(x\in A \And x\notin B).
A \ne B	Prove (A\not\subseteq B) \Or (B\not\subseteq A). ie, there is an x\in A where x\notin B or      there is an x\in B where x\notin A.
A \ne \emptyset	Prove (\exists x)(x\in A).
x\notin A \cup B	Prove (x\notin A) \And (x\notin B).
x\notin A \cap B	Prove (x\notin A) \Or (x\notin B).
x\notin A - B	Prove (x\notin A) \Or (x\in B).

To prove P by contradiction: "Assume P is false" then show that you arrive at a contradition. To prove (\forall n\in\N)(P(n)) by Mathematical Induction, show the following: P(1) is true For all k\in\N, if P(k) is true then P(k+1) is true.

Math 99 (Winter 2005) web pages Created: 01 Feb 2005 Last modified: Feb 1, 2005 8:38:22 PM Comments to: dpvc@union.edu

Quiz 1 Review Selected Course Notes