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
1. P ∨ Q
2. ~P
3. Therefore Q
1. P
2. Therefore P ∨ Q
1. P → Q
2. R → S
3. P ∨ R
4. Therefore Q ∨ S
[/column]Simplification:
1. P ∧ Q
2. Therefore P
1. P → Q
2. Q → R
3. Therefore P → R
1. P
2. Q
3. Therefore P ∧ Q
1. P → Q
2. ~Q
3. Therefore ~P
[end_columns]
Substitution Rules
[column width=”45%” padding=”5%”]Duplication:
P = P ∧ P
P = P ∨ P
P ∧ Q = Q ∧ P
P ∨ Q = Q ∨ P
P → Q = ~P ∨ Q
P ≡ Q = (P → Q) ∧ (Q → P)
P → Q = ~Q → P
[/column]DeMorgan’s:
~(P ∨ Q) = (~P ∧ ~Q)
~(P ∧ Q) = (~P ∨ ~Q)
((P ∧ Q) ∧ R) = (P ∧ (Q ∧ R))
((P ∨ Q) ∨ R) = (P ∨ (Q ∨ R))
(P ∨ (Q and; R)) = (P ∨ Q) ∧ (P ∨ R)
(P ∧ (Q ∨ R)) = (P ∧ Q) ∨ (P ∧ R)
((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!