Tutorial SL07: The Full Truth-Table Method
By: Jonathan Chan
SL07.1 The full truth-table method
In this tutorial we study how to make use of full truth-table method to check the validity of a
sequent in SL. Consider this valid sequent:
P, (P→Q) Q
To prove that it is valid, we draw a table where the top row contains all the different sentence
letters in the argument, followed by the premises, and then the conclusion:
Then using the same method as in drawing complex truth-tables, we list all the possible
assignments of truth-values to the sentence letters on the left. In our particular example, since
there are only two sentence letters, there should be 4 assignments :
P Q P (P→Q) Q
T T
T F
F T
F F
The next step is to draw the truth-table for each of the premise and also the conclusion. First, the
first premise :
P Q P
T T
T
T F
T
F T
F
F F
F
(P→Q) Q
Then the second premise:
P Q P (P→Q) Q
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T T
T
T
T F
T
F
F T
F
T
F F
F
T
Finally for the conclusion:
P Q P (P→Q) Q
T T
T
T
T
T F
T
F
F
F T
F
T
T
F F
F
T
F
In the completed truth-table, the first two cells in each row give us the assignment of truthvalues, and the next three cells tell us the truth-values of the premises and the conclusion under
each of the assignment. If an argument is valid, then every assignment where the premises are all
true is also an assignment where the conclusion is true. It so happens that there is only one
assignment (the first row) where both premises are true. We can see from the last cell of the row
that the conclusion is also true under such an assignment. So this argument has been shown to be
valid.
SL07.2 Exercises
Question 1 - Note that in the table above the conclusion is false in the second and the forth row.
Why can't we say that they show that the argument is invalid?
SL07.3 More examples
Remember that “(P→Q), ~P, therefore ~Q” is invalid. Verify for yourself that this is the truthtable for the sequent :
P Q (P→Q) ~P ~Q
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T T
T
F
F
T F
F
F
T
F T
T
T
F
F F
T
T
T
Check for yourself that the truth-table is done correctly. A sequent is invalid if there is at least
one assignment where the premises are true and the conclusion is false. The third row of the table
shows that this is indeed the case. When “P” is false and “Q” is true, the two premises are true
but the conclusion is false. So the argument is not valid. We call such an assignment an
invalidating assignment (a counterexample) for the sequent.
So far we have only looked at sequents where each WFF contains at most one sentential
connective. Now let us look at sequents with more complex WFFs, such as this one here :
(~P∨Q), ~(Q→P) (Q↔~P)
Again we draw a truth-table for the premises and the conclusion :
P Q (~P∨Q) ~(Q→P) (Q↔~P)
T T
FT T T
FTTT
T F FT
T F
FT F F
FFTT
F T FT
F T
TF T T
TTFF
T T TF
F F
TF T F
FFTF
F F TF
To help us calculate the truth-values of the WFFs under each assignment, we use the full truthtable method to write down the truth-values of the sentence letters first, and then work out the
truth-values of the whole WFFs step by step. The truth-values of the complete WFFs under each
assignment is written beneath the main operator of the WFFs. As you can see, there is no
assignment where the premises are true and the conclusion is false. So the sequent is valid.
SL07.4 Exercises
Question 1
Examine this table and answer the questions:
P Q ~Q (P↔Q) ~P
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T T
F
T
F
T F
T
F
F
F T
F
T
F
F F
T
T
T
Which sequent is being tested for validity in this table?
Is the sequent valid according to this table?
There are two mistakes made in this table. Can you identify them?
Question 2 - For any three formula φ1, φ2 and φ3, if φ1
φ1 φ3?
φ2, and φ2
φ3, does it follow that
Question 3 - Use the full truth-table method to determine the validity of these sequents:
((P→Q)&R), (~Q∨R)
(P↔(Q↔R))
(((P&Q)&~S) & (~Q↔R)), ~(~R→S)
((P→Q)→(S→R))
Question 4 - For any two formula φ and ψ, if φ
ψ, does it follow that ψ
φ?
Question 5 - For any two formula φ and ψ, if φ does not entail ψ, does it follow that ψ entail φ?
Question 6 Confirm for yourself that the WFFs in each pair of WFFs below are logically
equivalent to each other :
(P→Q), (~Q→~P)
(P↔Q), (~P↔~Q)
~(PvQ), (~P&~Q)
~(P&Q), (~Pv~Q)
SL07.5 Some informal comments on decidability
One thing you might notice about the full truth-table method is that it can help us determine the
validity of any sequent in SL. A program can be written that, given any finite sequent in SL as
input, after a finite number of processing steps, produces an output "Yes" if the sequent is valid,
or an output "No" if it is not. Of course, the computer would need to have a lot of memory if the
sequent is a long one, but in principle it can be done. This is roughly what logicians mean when
they say that validity in SL is decidable. Intuitively, it is a matter of whether there is an algorithm
or computer program that can come up with a proof of either the validity or the invalidity of a
sequent.
What is interesting and perhaps surprising is that decidability no longer obtains when we are
dealing with some more powerful systems of logic. One of the most important discoveries in
modern logic is that mathematics is undecidable. In particular, Gödel's first incompleteness
theorem says that there any consistent systems of mathematics would have to include statements
that can neither be proved or disproved. In other words, it is impossible for there to be a
computer program that can tell whether these statements are true.
Jonathan Chan