How do you construct a truth table?
- List component in increasing complexity
- Determine the amount of rows by 2 ^ # of atomic sentences
- Alternate truth table values, by 4, then 2, then within the rows
The Law of Double Negation (example and abbreviation)
DN. ~~x = x
De Morgan’s Law (ex. and ab.)
DM. 1. ~(X&Y) = ~Xv~Y
2. ~(XvY) = ~X&~Y
NOTICE THE SWITCH FROM CONJUNCTION TO DISJUNCTION, and vice versa
The Distributive Law (ex. and ab.)
D. For any 3 sentences, X,Y, and Z, X&(YvZ) = (X&Y)v(X&Z)
Pay attention to the connectives!!
The Law of Transitivity (ex. and ab.)
TLE. If X=Y and Y=Z, then X=Z.
Communicative Law (ex. and ab.)
CM. X&Y=Y&X
Disjuncts and conjuncts.
Associative Law (ex. and ab.)
A. X&(Y&Z)=(X&Y)&Z
Law of Redundancy (ex. and ab.)
RD. X&X=X
Logical Truth Example
Mv~M
Contradiction Example
M&~M
Law of Logically True Conjunct (ex. and ab.)
LTC. X is any sentence and Y is a LT, then X&Y = X
Law of Contradictory Disjunct (ex. and ab.)
LCD. X is any sentence and Y is ant C, then XvY = X
A conjunction is always a contradiction when…
one of its conjuncts is a contradiction.
A disjunction is always a logical truth when…
one of its disjuncts in a logical truth.
How to determine Disjunctive Normal Form
- Draw the truth table
- In every case (row), put conjunction between every true column.
- Put disjuncts between the newly created compounds with conjuncts.
Define Expressive Completeness
When a connective, or set of them can represent every truth function.
Disjunction, conjunction, and negation, taken together, are Expressively Complete.
How to determine validity from a truth table
- Look when the conclusion is true
- In those places, if every premise is true, circle it
- If there are any counterexamples, or those rows where the conclusion is true but not ALL the premises are, then invalid argument
- If it’s made it this far, it’s valid
Law of Contraposition (ex. and ab.)
CP. X->Y = ~Y->~X
Law of the Conditional (ex. and ab.)
C. X->Y = ~(X&Y) and (~XvY)
Law of the BiConditional (ex. and ab.)
B. A biconditional is the same as (X->Y)&(Y->X)
Explain Modus Tollens (denying the consequent) give abbreviation
DC. Assume the antecedent and show the contradiction to arrive at the conclusion
Explain Reductio Ad Adsurdum (ab.)
RD. From X, arriving at Y&~Y, then X is actually ~X
Explain Argument by Cases. (ab.)
AC. If a disjunction, and the disjuncts both have the same consequent in a conditional, you prove the consequent as true.
Still must show the mini-proof!
