cse 260 exam 1

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1
Identity Laws
p∧T≡ p
p∨F≡ p
2
prepositions
only true or false (not both)
3
p∧T≡ p
p∨F≡ p

Prepositions:
4
Only true or false (not both)


Converse:
5
q then p


Contrapositive:
6
¬q →¬p


Inverse:
7
¬p →¬q


Contingency:
8
Neither contradiction nor tautology


Domination Laws:
9
p∨T≡ p
p∧F≡ p

Idempotent:
10
p∨p ≡ p
p∧p ≡ p

Commutative laws:
11
p∨q ≡ q ∨p
p∧q ≡ q ∧p

Distributive laws:
12
p∨(q ∧r)≡ (p∨q)∧(p∨r)
p∧(q ∨r)≡ (p∧q)∨(p∧r)

¬p:
13
Not p


p∧q:
14
p and q


p∨q:
15
p or q


p → q:
16
If p then q. Only false when p is true and q is false. p implies q


p⊕q:
17
Exclusive or. True only when exactly one of p or q is true


p ↔ q:
18
IFF. True when both have the same truth values


Contradiction:
19
Always false


Tautology:
20
Always true


Double Negation:
21
¬(¬p)≡ p


Associative laws:
22
(p∨q)∨r ≡ p∨(q ∨r)
(p∧q)∧r ≡ p∧(q ∧r)

De Morgan's laws:
23
¬(p∧q)≡¬p∨¬q
¬(p∨q)≡¬p∧¬q

Absorption laws:
24
p∨(p∧q)≡ p
p∧(p∨q)≡ p

Negation Laws:
25
p∨¬p ≡T
p∧¬p ≡F

Modus Ponens:
26
Assume p, p → q, therefore q (tautology)


Modus Tollens:
27
¬q, p → q, therefore ¬p


Hypothetical Syllogism:
28
p → q, q → r, therefore p → r


Disjunctive Syllogism:
29
p∨q, ¬p, therefore q


Addition:
30
p, therefore p∨q


Simplification:
31
p∧q, therefore p


Conjunction:
32
p, q, therefore p∧q


Resolution:
33
p∨q, ¬p∨r, therefore q ∨r
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