Rule Index

We’ve finished all of the rules but the two proof applications, so I thought I would do a quick rundown of how each rule works. No guest arguments today, this is going to be a lightning round of symbols and links to help you mind your p’s and q’s. 

Operators

[column width=”45%” padding=”5%”]∧ – Conjunction, “and”

∨ – Disjunction, “or”

→ – Material Conditional, “if-then”

[/column]~ – Negation, “not”

≡ – Biconditional, “if and only if”

[end_columns]

Rules of Inference

[column width=”45%” padding=”5%”]Modus Ponens:

1. P → Q
2. P
3. Therefore Q

Disjunctive Syllogism:

1. P ∨ Q
2. ~P
3. Therefore Q

Addition:

1. P
2. Therefore P ∨ Q

Dilemma:

1. P → Q
2. R → S
3. P ∨ R
4. Therefore Q ∨ S

[/column]Simplification:

1. P ∧ Q
2. Therefore P

Hypothetical Syllogism:

1. P → Q
2. Q → R
3. Therefore P → R

Conjunction:

1. P
2. Q
3. Therefore P ∧ Q

Modus Tollens:

1. P → Q
2. ~Q
3. Therefore ~P

[end_columns]

Substitution Rules

[column width=”45%” padding=”5%”]Duplication:

P = P ∧ P
P = P ∨  P

Commutation:

P ∧ Q = Q ∧ P
P ∨ Q = Q ∨ P

Conditional Exchange:

P → Q = ~P ∨ Q

Biconditional Exchange:

P ≡ Q = (P → Q) ∧ (Q → P)

Contraposition:

P → Q = ~Q → P

[/column]DeMorgan’s:

~(P ∨ Q) = (~P ∧ ~Q)
~(P ∧ Q) = (~P ∨  ~Q)

Association:

((P ∧ Q) ∧ R) = (P ∧ (Q ∧ R))
((P ∨  Q) ∨ R) = (P ∨ (Q ∨ R))

Distribution:

(P ∨ (Q and; R)) = (P ∨ Q) ∧ (P ∨ R)
(P ∧ (Q ∨ R)) = (P ∧ Q) ∨ (P ∧ R)

Exportation:

((P ∧ Q) → R) = (P → (Q → R))[end_columns]

And that’s it, all of the logical rules we’ve learned so far, in a compact form. Thanks for following along!