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Summary of What to Prove:

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 $P(x)$ for this $x$.
$(\exists x)(P(x))$ "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:

  1. $P(1)$ is true
  2. For all $k\in\N$, if $P(k)$ is true then $P(k+1)$ is true.

[HOME] Math 199 (Fall 2007) web pages
Created: 30 Aug 2007
Last modified: 30 Aug 2007 00:00:00
Comments to: dpvc@union.edu
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