For Wednesday, read Chapter 3, section 4.
Nongraded Homework: Problems at the end
of section 4, set I only; Power of Logic web
tutor, 7.4, A, B, and C
Graded HW#3 is due at the beginning of
class on Friday.
~ [ ~ ( ~ A → B) ↔ (A & ~ D)]
Let’s say that A and B are false and D is true.
Using truth-tables to classify formulae:
First, construct a truth-table:
1. determine the number of rows needed; this
is 2n, where n = the number of distinct
sentence letters in the formula;
2. write the formula horizontally across the
top of the table
3. To the left of the formula, list (horizontally, left to
right) the distinct statement letters in the order in
which they appear in the formula; put alternating
Ts and ^s beneath the sentence letter to the far
right (closest to the premises); moving to the left,
continue to fill in Ts and ^s, doubling the size of
the groups of Ts and ^s for each new letter you
come to. The last letter to the left should have
2n/2 Ts followed by 2n/2 ^s beneath it.
Calculate truth-values of the entire formula on each
row of the truth-table.
Tautology: True on every row of its truth-table (Ts
all the way down underneath the main connective)
Contradiction: False on every row of its truth-table
(^s all the way down underneath the main
connective)
Contingent Statement: True on at least one row
and false on at least one row of its truth-table (at
least one T and at least one ^ underneath the main
connective).
A B
T T
T^
^T
^^
(A ↔ B) → ( ~ A & ~ B)
T
^ ^ ^^
^
T ^ ^T
^
T T ^^
T
T T TT
*
Answer: Contingent
A B
T T
T^
^T
^^
A → [~ (A & B) → ~ B]
T ^ T
T ^
T T ^
T T
T
T
*
Answer: Tautology
F G
T T
T^
^T
^^
~ (F → G) & G
^ T
^
^
^
T
^
^
*
Answer: Contradiction
Using truth-tables to test for logical
equivalence:
Make a truth-table that includes both
formulae. If their truth-values match on each
row, they are logically equivalent; if there is
even one line where the truth-values of the
two statements (as a whole) don’t match, the
statements are not logically equivalent.
D G
T T
T^
^T
^^
~ D v ~ G ~ (D & G)
^ ^^
^
T
^ TT
T
^
T T^
T
^
T TT
T
^
*
*
On each line, the value of the m.o.’s match
each other. So, these two statements are
logically equivalent.
Using Truth-tables to test arguments for validity.
Place the entire argument on a truth-table: list the
premises from left to right, separated by commas;
put the conclusion on the far right after a ‘\’ .
Fill out the table. If there is at least one row where
all premises are true and the conclusion false, then
the argument is invalid;
if there is no row with all true premises and a false
conclusion, the argument is valid.
Z → (S v G), Z & G \ S
T
T
T
T
T
T
^
T
T
T
T
^ *****
^
^
^
^
T
T
^
T
T
T
^
T
T
T
^
^
T
^
^
^
*
*
*
Answer: Invalid
ZSG
TTT
TT^
T^T
T^^
^TT
^T^
^^T
^^^
Why does it work?
Each row on the truth-table represents a relevant
possibility (an interpretation), and taken together
the rows represent all of the relevant possibilities
(all possible interpretations).
So, if there is a row on the truth-table with all true
premises and a false conclusion, then it is
possible to have all true premises and a false
conclusion—thus the argument is invalid;
and if there is no row with all true premises and a
false conclusion, it is impossible to have all true
premises and a false conclusion, and by
definition, the argument is valid.
PQS
TTT
TT^
T^T
T^^
^TT
^T^
^^T
^^^
P → Q, S → Q, ~ Q \ ~ P & ~ S
T
T
^
^ ^^
T
T
^
^ ^T
^
^
T
^ ^^
^
T
T
^ ^T
T
T
^
T ^^
T
T
^
T TT
T
^
T
T ^^
T
T
T
T TT
Answer? Valid. There is no line showing all
true premises and a false conclusion. Thus,
it is impossible for the argument to have all
true premises and a false conclusion. So,
the reasoning can’t go wrong. Whenever the
premises are all true, the conclusion is as
well.
Remember, when an argument is valid, there
is no particular line that proves validity.
ABG
A → (B & G), ~ B \ A  ~ B
TTT
T
^
TT^
^
^
T^T
^
^
T^^
^
^
^TT
T
^
^T^
T
^
^^T
T
T
^^T
FFF
Answer: Invalid, proven by line seven