Chapter 11 Complete truth tables
So far, we have used symbolization keys to assign truth values to TFL sentences indirectly. For example, we might say that the TFL sentence ‘$B$’ is to be true iff Big Ben is in London. Since Big Ben is in London, this symbolisation would make ‘$B$’ true. But we can also assign truth values directly. We can simply stipulate that ‘$B$’ is to be true, or stipulate that it is to be false. Such stipulations are called valuations:
A valuation is any assignment of truth values to particular sentence letters of TFL.
The power of truth tables lies in the following. Each row of a truth table represents a possible valuation. The complete truth table represents all possible valuations. And the truth table provides us with a means to calculate the truth value of complex sentences, on each possible valuation. But all of this is easiest to explain by example.
11.1 A worked example
Consider the sentence ‘$(H\wedge I)\to H$’. There are four possible ways to assign True and False to the sentence letter ‘$H$’ and ‘$I$’—four valuations—which we can represent as follows:
$H$  $I$  $(H$  $\wedge $  $I)$  $\to $  $H$ 
T  T  
T  F  
F  T  
F  F 
To calculate the truth value of the entire sentence ‘$(H\wedge I)\to H$’, we first copy the truth values for the sentence letters and write them underneath the letters in the sentence:
$H$  $I$  $(H$  $\wedge $  $I)$  $\to $  $H$ 
T  T  T  T  T  
T  F  T  F  T  
F  T  F  T  F  
F  F  F  F  F 
Now consider the subsentence ‘$(H\wedge I)$’. This is a conjunction, $(\mathcal{A}\wedge \mathcal{B})$, with ‘$H$’ as $\mathcal{A}$ and with ‘$I$’ as $\mathcal{B}$. The characteristic truth table for conjunction gives the truth conditions for any sentence of the form $(\mathcal{A}\wedge \mathcal{B})$, whatever $\mathcal{A}$ and $\mathcal{B}$ might be. It represents the point that a conjunction is true iff both conjuncts are true. In this case, our conjuncts are just ‘$H$’ and ‘$I$’. They are both true on (and only on) the first line of the truth table. Accordingly, we can calculate the truth value of the conjunction on all four rows.
$\mathcal{A}$  $\wedge $  $\mathcal{B}$  
$H$  $I$  $(H$  $\wedge $  $I)$  $\to $  $H$ 
T  T  T  T  T  T  
T  F  T  F  F  T  
F  T  F  F  T  F  
F  F  F  F  F  F 
Now, the entire sentence that we are dealing with is a conditional, $\mathcal{A}\to \mathcal{B}$, with ‘$(H\wedge I)$’ as $\mathcal{A}$ and with ‘$H$’ as $\mathcal{B}$. On the second row, for example, ‘$(H\wedge I)$’ is false and ‘$H$’ is true. Since a conditional is true when the antecedent is false, we write a ‘T’ in the second row underneath the conditional symbol. We continue for the other three rows and get this:
$\mathcal{A}$  $\to $  $\mathcal{B}$  
$H$  $I$  $(H$  $\wedge $  $I)$  $\to $  $H$ 
T  T  T  T  T  
T  F  F  T  T  
F  T  F  T  F  
F  F  F  T  F 
The conditional is the main logical operator of the sentence, so the column of ‘T’s underneath the conditional tells us that the sentence ‘$(H\wedge I)\to H$’ is true regardless of the truth values of ‘$H$’ and ‘$I$’. They can be true or false in any combination, and the compound sentence still comes out true. Since we have considered all four possible assignments of truth and falsity to ‘$H$’ and ‘$I$’—since, that is, we have considered all the different valuations—we can say that ‘$(H\wedge I)\to H$’ is true on every valuation.
In this example, we have not repeated all of the entries in every column in every successive table. When actually writing truth tables on paper, however, it is impractical to erase whole columns or rewrite the whole table for every step. Although it is more crowded, the truth table can be written in this way:
$H$  $I$  $(H$  $\wedge $  $I)$  $\to $  $H$ 
T  T  T  T  T  T  T 
T  F  T  F  F  T  T 
F  T  F  F  T  T  F 
F  F  F  F  F  T  F 
Most of the columns underneath the sentence are only there for bookkeeping purposes. The column that matters most is the column underneath the main logical operator for the sentence, since this tells you the truth value of the entire sentence. We have emphasized this, by putting this column in bold. When you work through truth tables yourself, you should similarly emphasize it (perhaps by highlighting).
11.2 Building complete truth tables
A complete truth table has a line for every possible assignment of True and False to the relevant sentence letters. Each line represents a valuation, and a complete truth table has a line for all the different valuations.
The size of the complete truth table depends on the number of different sentence letters in the table. A sentence that contains only one sentence letter requires only two rows, as in the characteristic truth table for negation. This is true even if the same letter is repeated many times, as in the sentence ‘$[(C\leftrightarrow C)\to C]\wedge \mathrm{\neg}(C\to C)$’. The complete truth table requires only two lines because there are only two possibilities: ‘$C$’ can be true or it can be false. The truth table for this sentence looks like this:
$C$  $[($  $C$  $\leftrightarrow $  $C$  $)$  $\to $  $C$  $]$  $\wedge $  $\mathrm{\neg}$  $($  $C$  $\to $  $C$  $)$ 
T  T  T  T  T  T  F  F  T  T  T  
F  F  T  F  F  F  F  F  F  T  F 
Looking at the column underneath the main logical operator, we see that the sentence is false on both rows of the table; i.e., the sentence is false regardless of whether ‘$C$’ is true or false. It is false on every valuation.
There will be four lines in the complete truth table for a sentence containing two sentence letters, as in the characteristic truth tables, or the truth table for ‘$(H\wedge I)\to H$’.
There will be eight lines in the complete truth table for a sentence containing three sentence letters, e.g.:
$M$  $N$  $P$  $M$  $\wedge $  $(N$  $\vee $  $P)$ 
T  T  T  T  T  T  T  T 
T  T  F  T  T  T  T  F 
T  F  T  T  T  F  T  T 
T  F  F  T  F  F  F  F 
F  T  T  F  F  T  T  T 
F  T  F  F  F  T  T  F 
F  F  T  F  F  F  T  T 
F  F  F  F  F  F  F  F 
From this table, we know that the sentence ‘$M\wedge (N\vee P)$’ can be true or false, depending on the truth values of ‘$M$’, ‘$N$’, and ‘$P$’.
A complete truth table for a sentence that contains four different sentence letters requires 16 lines. Five letters, 32 lines. Six letters, 64 lines. And so on. To be perfectly general: If a complete truth table has $n$ different sentence letters, then it must have ${2}^{n}$ lines.
In order to fill in the columns of a complete truth table, begin with the rightmost sentence letter and alternate between ‘T’ and ‘F’. In the next column to the left, write two ‘T’s, write two ‘F’s, and repeat. For the third sentence letter, write four ‘T’s followed by four ‘F’s. This yields an eight line truth table like the one above. For a 16 line truth table, the next column of sentence letters should have eight ‘T’s followed by eight ‘F’s. For a 32 line table, the next column would have 16 ‘T’s followed by 16 ‘F’s, and so on.
11.3 More about brackets
Consider these two sentences:
$((A\wedge B)\wedge C)$  
$(A\wedge (B\wedge C))$ 
These are truth functionally equivalent. Consequently, it will never make any difference from the perspective of truth value—which is all that TFL cares about (see chapter 10)—which of the two sentences we assert (or deny). Even though the order of the brackets does not matter as to their truth, we should not just drop them. The expression
$A\wedge B\wedge C$ 
is ambiguous between the two sentences above. The same observation holds for disjunctions. The following sentences are logically equivalent:
$((A\vee B)\vee C)$  
$(A\vee (B\vee C))$ 
But we should not simply write:
$A\vee B\vee C$ 
In fact, it is a specific fact about the characteristic truth table of $\vee $ and $\wedge $ that guarantees that any two conjunctions (or disjunctions) of the same sentences are truth functionally equivalent, however you place the brackets. This is only true of conjunctions and disjunctions, however. The following two sentences have different truth tables:
$((A\to B)\to C)$  
$(A\to (B\to C))$ 
So if we were to write:
$A\to B\to C$ 
it would be dangerously ambiguous. Leaving out brackets in this case would be disastrous. Equally, these sentences have different truth tables:
$((A\vee B)\wedge C)$  
$(A\vee (B\wedge C))$ 
So if we were to write:
$A\vee B\wedge C$ 
it would be dangerously ambiguous. Never write this. The moral is: never drop brackets (except the outermost ones).
Practice exercises
A. Offer complete truth tables for each of the following:

1.
$A\to A$

2.
$C\to \mathrm{\neg}C$

3.
$(A\leftrightarrow B)\leftrightarrow \mathrm{\neg}(A\leftrightarrow \mathrm{\neg}B)$

4.
$(A\to B)\vee (B\to A)$

5.
$(A\wedge B)\to (B\vee A)$

6.
$\mathrm{\neg}(A\vee B)\leftrightarrow (\mathrm{\neg}A\wedge \mathrm{\neg}B)$

7.
$\left[(A\wedge B)\wedge \mathrm{\neg}(A\wedge B)\right]\wedge C$

8.
$[(A\wedge B)\wedge C]\to B$

9.
$\mathrm{\neg}\left[(C\vee A)\vee B\right]$
B. Check all the claims made in section 11.3, i.e., show that:

1.
‘$((A\wedge B)\wedge C)$’ and ‘$(A\wedge (B\wedge C))$’ have the same truth table

2.
‘$((A\vee B)\vee C)$’ and ‘$(A\vee (B\vee C))$’ have the same truth table

3.
‘$((A\vee B)\wedge C)$’ and ‘$(A\vee (B\wedge C))$’ do not have the same truth table

4.
‘$((A\to B)\to C)$’ and ‘$(A\to (B\to C))$’ do not have the same truth table
Also, check whether:

5.
‘$((A\leftrightarrow B)\leftrightarrow C)$’ and ‘$(A\leftrightarrow (B\leftrightarrow C))$’ have the same truth table
C. Write complete truth tables for the following sentences and mark the column that represents the possible truth values for the whole sentence.

1.
$\mathrm{\neg}(S\leftrightarrow (P\to S))$

2.
$\mathrm{\neg}[(X\wedge Y)\vee (X\vee Y)]$

3.
$(A\to B)\leftrightarrow (\mathrm{\neg}B\leftrightarrow \mathrm{\neg}A)$

4.
$[C\leftrightarrow (D\vee E)]\wedge \mathrm{\neg}C$

5.
$\mathrm{\neg}(G\wedge (B\wedge H))\leftrightarrow (G\vee (B\vee H))$
D. Write complete truth tables for the following sentences and mark the column that represents the possible truth values for the whole sentence.

1.
$(D\wedge \mathrm{\neg}D)\to G$

2.
$(\mathrm{\neg}P\vee \mathrm{\neg}M)\leftrightarrow M$

3.
$\mathrm{\neg}\mathrm{\neg}(\mathrm{\neg}A\wedge \mathrm{\neg}B)$

4.
$[(D\wedge R)\to I]\to \mathrm{\neg}(D\vee R)$

5.
$\mathrm{\neg}[(D\leftrightarrow O)\leftrightarrow A]\to (\mathrm{\neg}D\wedge O)$
If you want additional practice, you can construct truth tables for any of the sentences and arguments in the exercises for the previous chapter.