7.4 Abbreviated Truth Tables
The full truth table method of Section 7.3 is extremely cumbersome.
For example, an argument with only four statement letters requires
a truth table with 24 = 32 rows. One with five requires a truth table
with 25 = 64 rows. Obviously, truth tables of these sizes are simply
impractical to construct. Abbreviated truth tables provide a much
more efficient method for determining validity.
The method
The key insight behind the method
If we can construct just one row of a truth table for an argument that makes the premises true and the conclusion false,
then we will have shown the argument to be invalid. If we fail
at such an attempt, we will have shown the argument to be
valid.
Recall the argument from the lecture for 7.3:
Abortion is permissible only if fetuses are not innocent human beings
or it is not always wrong to kill innocent human beings. But it is always
wrong to kill innocent human beings. So abortion is not permissible. (A:
Abortion is permissible; B: Fetuses are innocent human beings; W: It is
always wrong to kill innocent human beings.)
We symbolized this sentence as follows:
[A → (∼B ∨ ∼W)]. W ∴ ∼A
The method applied to an invalid argument
1. Write down the symbolized argument thus (note that we’ve dispensed with periods; they just get in the way):
A → (∼B ∨ ∼W),
W ∴ ∼A
2. Hypothesize that the premises are true and the conclusion false:
A → (∼B ∨ ∼W),
T
W ∴ ∼A
T
F
3. Calculate the immediate consequences of this hypothesis:
◦ Copy the truth value assigned to W to its other occurrence:
A → (∼B ∨ ∼W),
T
T
W ∴ ∼A
T
F
◦ Calculate truth values of the compound statements whenever you know the
truth values of their component parts. Thus, we can calculate that ∼W is false
in virtue of our assumption that W is true:
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A → (∼B ∨ ∼W),
T
FT
W ∴ ∼A
T
F
◦ Calculate truth values of the component parts of compound statements whenever you know the truth values of the compound statements (and copy any truth
values assigned to statement letters to all other occurrences). Thus, we can
calculate that A must be true given that ∼A is false:
A → (∼B ∨ ∼W),
T T
FT
W ∴ ∼A
T
FT
Furthermore, since (i) we’ve hypothesized that [A → (∼B ∨ ∼W)] is true and
(ii) we have calculated that the antecedent A of [A → (∼B ∨ ∼W)] must be true
(given our initial hypothesis about ∼A), it now follows that (∼B ∨ ∼W) must be
true as well, lest we falsify our hypothesis that [A → (∼B ∨ ∼W)] is true. (Recall
that a conditional is false if its antecedent is true and its consequent is false.)
Hence:
A → (∼B ∨ ∼W),
T T
T FT
W ∴ ∼A
T
FT
But now that we have deduced that the disjunction (∼B ∨ ∼W) must be true
(given our initial hypotheses), we know that, because its right disjunct ∼W is
false, its left disjunct ∼B must be true — for a disjunction is true if and only if at
least one of its disjuncts is. Thus:
A → (∼B ∨ ∼W),
T T T T FT
W ∴ ∼A
T
FT
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And this, of course, enables us now to deduce that B is false, which completes
the row:
A → (∼B ∨ ∼W),
T T TF T FT
W ∴ ∼A
T
FT
So we have identified a row that makes the premises of our argument true and
the conclusion false, so we have thereby demonstrated that the argument is
invalid when A is true, B is false, and W is true. We complete the abbreviated
truth table by recording this invalidating truth value assignment into the table:
A B W A → (∼B ∨ ∼W),
T F T T T TF T FT
W ∴ ∼A
T
FT
Much shorter than the full truth table! To remind you:
A
T
T
T
T
F
F
F
F
B
T
T
F
F
T
T
F
F
W A → (∼B
T
F
F
T
F
F
T
T
T
F
T
T
T
T
F
T
F
F
T
T
T
F
T T
∨ ∼W),
F
F
T
T
T
F
T
T
F
F
T
T
T
F
W ∴ ∼A
T
F
F
F
T
F
F
F
T
T
F
T
T
T
T
F
T
T
Note that row 3 is exactly the row that we just constructed using the abbreviated
truth table method.
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The method applied to a valid argument
What happens if the argument in question is valid ? We demonstrate with a
further example. We’ll cut right to the chase with a symbolized argument without
worrying about the English argument it symbolizes.
We begin with the usual hypothesis that the premises are true and the conclusion
false:
W ∨ J,
T
(W → Z) ∨ (J → Z),
T
∼Z ∴ ∼(W • J)
T
F
Since the conclusion ∼(W • J) is false, its component statement (W • J) must
be true:
W ∨ J,
T
(W → Z) ∨ (J → Z),
T
∼Z ∴ ∼(W • J)
T
F
T
From the truth table schema for •, the only way for (W • J) to be true is if both
W and J are true. So we record this information below the conclusion, and copy
it into the rest of the table:
W ∨ J,
T T T
(W → Z) ∨ (J → Z),
T
T T
∼Z ∴ ∼(W • J)
T
F T T T
So far so good. Now, from the fact that ∼Z is true, we infer that Z is false, and
record this information into the table:
W ∨ J,
T T T
(W → Z) ∨ (J → Z),
T
F T T
F
∼Z ∴ ∼(W • J)
TF
F T T T
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This then enables us to fill in the truth values for the two disjuncts in the second
premise in accordance with the truth table schema for conditions:
W ∨ J,
T T T
(W → Z) ∨ (J → Z),
T F F T T F F
∼Z ∴ ∼(W • J)
TF
F T T T
But now there is a problem: our hypothesis was that the three premises were all
true. But the second premise is a disjunction with two false disjuncts. Hence,
it is itself false. But this conflicts with our initial hypothesis. Hence, we have
shown that it is not possible after all for the premises of the argument to be
true and the conclusion false. We indicate the contradiction that arises when we
assume otherwise by putting ‘T/F’ underneath the main logical operator of the
problematic premise (all of them, if there are more than one):
W ∨ J,
T T T
(W → Z) ∨ (J → Z),
T F F T/F T F F
∼Z ∴ ∼(W • J)
TF
F T T T
When there are several ways the conclusion can be false
If the conclusion can be made false in more than one way, then every way must
be tried until either an invalidating assignment is found or all the ways of making
the conclusion false are exhausted. Again, we start with the usual hypothesis:
∼B → ∼A,
T
C → D,
T
A • C ∴ B • D
T
F
But note that, by the truth table schema for •, there are three ways to make the
conclusion false. So we simply pick one of the three to start with, viz., the one
where both B and D are false:
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∼B → ∼A,
T
C → D,
T
A • C ∴ B • D
T
F F F
Copying these to the rest of the row and calculating truth values we get:
∼B → ∼A,
TF T
C → D,
T F
A • C ∴ B • D
T
F F F
Because of the truth table schema for → and our hypothesis that (C → D) is
true, it follows that C has to be false as well:
∼B → ∼A,
TF T
C → D,
F T F
A • C ∴ B • D
T
F F F
Copying this value to (A • C) we get
∼B → ∼A,
T F T
C → D,
F T F
A • C ∴ B • D
T F
F F F
But now we see that there will be no way to make (A • C) true, since one of its
conjuncts is false. Consequently, this particular way of making the conclusion
false fails to yield an invalidating assignment, and we so indicate:
∼B → ∼A,
TF T
C → D,
F T F
A • C ∴ B • D
T/F F
F F F
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Note: The Web Tutor will require you to complete the row. To do so, just assign
an arbitrary value to A (it won’t matter because (A • C) will be false either way
just because C is); ‘F’ is easiest in this case:
∼B → ∼A,
TF T TF
C → D,
F T F
A • C ∴ B • D
F T/F F
F F F
Now we need to try the two remaining ways of making the conclusion false; we
will (arbitrarily) try the one where B is true and D is false first:
∼B → ∼A,
TF T TF
T
C → D,
F T F
T
A • C ∴ B • D
F T/F F
F F F
T
T F F
But, when calculated out, this possibility fares no better than the first:
∼B → ∼A,
TF T TF
FT T FT
C → D,
F T F
T T/F F
A • C ∴ B • D
F T/F F
F F F
T T T
T F F
So we must try the final case, where B is false and D is true:
∼B → ∼A,
TF T TF
FT T FT
T
C → D,
F T F
T T/F F
T
A • C ∴ B • D
F T/F F
F F F
T T T
T F F
T
F F T
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But, when calculated out, this possibility too fails to yield an invalidating assignment:
∼B
TF
FT
TF
→
T
T
T
∼A,
TF
FT
TF
C → D,
F T F
T T/F F
T T T
A • C ∴ B • D
F T/F F
F F F
T T T
T F F
F T/F T
F F T
Only now that we have exhausted the possible ways of making the conclusion
false are we permitted to say that the argument is valid; for on all the possible ways of making the conclusion false we found we were unable to make the
premises true. If, however, any of these ways of making the conclusion false
had yielded a row with true premises, the argument would have been shown to
be invalid. To emphasize: to show invalidity, all it takes is one row, one truth
value assignment to the statement letters, on which the premises are true and
the conclusion false.
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