Table 1.3.1 - Epp Textbook p. 40

Modus Ponens	Modus Tollens		Disjunctive	$p \vee q$	$p \vee q$
$p \rightarrow q$	$p \rightarrow q$		Syllogism	$\sim q$	$\sim p$
$p$	$\sim q$			Therefore $p$	Therefore $q$
Therefore $q$	Therefore $\sim p$
Conjunctive	$p$		Hypothetical	$p \rightarrow q$
Addition	$q$		Syllogism	$q \rightarrow r$
	Therefore $p \wedge q$			Therefore $p \rightarrow r$
Disjunctive	$p$	$q$	Dilemma:	$p \vee q$
Addition	Therefore $p \vee q$	Therefore $p \vee q$	Proof by	$p \rightarrow r$
			Division	$q \rightarrow r$
			into Cases	Therefore $r$
Conjunctive	$p \wedge q$	$p \wedge q$	Rule of	$\sim p \rightarrow c$
Simplification	Therefore $p$	Therefore $q$	Contradiction	Therefore $p$
Closing C.W.	$\vert p$ Assumed		Closing C.W.	$\vert p$ Assumed
without	$\vert q$ derived		with	$\vert x \wedge \sim x$ derived
contradiction	Therefore $p \rightarrow q$		contradiction	Therefore $\sim p$

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