For Wednesday, read Chapter 3, section 3
(pp. 58-62).
Nongraded Homework: Exercises on p. 62
and Power of Logic, 7.3, A and B.
The logical connectives are truth-functions:
for each possible combination of truth-values,
a determinate output is specified.
Represented graphically for the ampersand:
p q
p&q
T T
T
T ^
^
^ T
^
^ ^
^
Read the lines horizontally, left to right; line 1
says, “When both conjuncts are true, the
entire ampersand-statement is true”.
In summary, an ampersand-statement is true
when both sides are true; false otherwise.
A wedge-statement is false when both sides
are false; it’s true otherwise.
A tilde “reverses” the truth-value of the entire
statement to which it applies.
A double-arrow statement is true if and only if
the truth-values of its two components match.
An arrow statement is false when the left side
(the antecedent) is true and the right side
(the consequent) is false; it’s true otherwise.
From this knowledge, we can calculate the
truth-value of any wff if we know the truthvalue of its component statement letters.
Example:
(A v ~ B)  C
Say that A and B are false, and C is true.
Order of work:
1. Put truth-values under all capital letters
2. Apply tildes that are to the immediate left of
capital letters
3. Apply tildes that are to the immediate left of tildes
that are to the immediate left of capital letters
(repeat as necessary)
4. Calculate truth-values for binary connectives
inside the smallest sets of parentheses (those that
contain no other parentheses)
5. Calculate truth-values for tildes that have only the
smallest sets of parentheses in their scope
6. Repeat steps 4 and 5 for larger and larger
parenthetical groupings until only the main
connective has no truth-value beneath it; calculate
its truth-value and you’re done—that’s the truthvalue of the entire compound statement.
~ [ ~ ( ~ 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.