Chapter 12 Semantic concepts
In the previous chapter, we introduced the idea of a valuation and showed how to determine the truth value of any TFL sentence, on any valuation, using a truth table. In this chapter, we will introduce some related ideas, and show how to use truth tables to test whether or not they apply.
12.1 Tautologies and contradictions
In chapter 3, we explained necessary truth and necessary falsity. Both notions have surrogates in TFL. We will start with a surrogate for necessary truth.
$\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}$ is a tautology iff it is true on every valuation.
We can use truth tables to decide whether a sentence is a tautology. If the sentence is true on every line of its complete truth table, then it is true on every valuation, so it is a tautology. In the example of chapter 11, ‘$(H\wedge I)\to H$’ is a tautology.
This is only, though, a surrogate for necessary truth. There are some necessary truths that we cannot adequately symbolize in TFL. One example is ‘$2+2=4$’. This must be true, but if we try to symbolize it in TFL, the best we can offer is a sentence letter, and no sentence letter is a tautology. Still, if we can adequately symbolize some English sentence using a TFL sentence which is a tautology, then that English sentence expresses a necessary truth.
We have a similar surrogate for necessary falsity:
$\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}$ is a contradiction (in TFL) iff it is false on every valuation.
We can use truth tables to decide whether a sentence is a contradiction. If the sentence is false on every line of its complete truth table, then it is false on every valuation, so it is a contradiction. In the example of chapter 11, ‘$[(C\leftrightarrow C)\to C]\wedge \mathrm{\neg}(C\to C)$’ is a contradiction.
12.2 Equivalence
Here is a similar useful notion:
$\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}$ and $\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{B}$}$ are equivalent (in TFL) iff, for every valuation, their truth values agree, i.e., if there is no valuation in which they have opposite truth values.
We have already made use of this notion, in effect, in section 11.3; the point was that ‘$(A\wedge B)\wedge C$’ and ‘$A\wedge (B\wedge C)$’ are equivalent. Again, it is easy to test for equivalence using truth tables. Consider the sentences ‘$\mathrm{\neg}(P\vee Q)$’ and ‘$\mathrm{\neg}P\wedge \mathrm{\neg}Q$’. Are they equivalent? To find out, we construct a truth table.
$P$  $Q$  $\mathrm{\neg}$  $(P$  $\vee $  $Q)$  $\mathrm{\neg}$  $P$  $\wedge $  $\mathrm{\neg}$  $Q$ 
T  T  F  T  T  T  F  T  F  F  T 
T  F  F  T  T  F  F  T  F  T  F 
F  T  F  F  T  T  T  F  F  F  T 
F  F  T  F  F  F  T  F  T  T  F 
Look at the columns for the main logical operators; negation for the first sentence, conjunction for the second. Go through the rows in the table one by one and compare the truth values in the columns for the main logical operators. On the first three rows, both are false. On the final row, both are true. Since they match on every row, the two sentences are equivalent.
12.3 Satisfiability
In chapter 3, we said that sentences are jointly possible iff it is possible for all of them to be true at once. We can offer a surrogate for this notion too:
${\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$1$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$2$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathrm{\dots}$}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$n$}}$ are jointly satisfiable (in TFL) iff there is some valuation which makes them all true.
Derivatively, sentences are jointly unsatisfiable iff no valuation makes them all true. Again, it is easy to test for joint satisfiability using truth tables.
12.4 Entailment and validity
The following idea is closely related to that of joint satisfiability:
The sentences ${\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$1$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$2$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathrm{\dots}$}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$n$}}$ entail (in TFL) the sentence $\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{C}$}$ iff no valuation of the relevant sentence letters makes all of ${\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$1$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$2$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathrm{\dots}$}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$n$}}$ true and $\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{C}$}$ false.
Again, it is easy to test this with a truth table. To check whether ‘$\mathrm{\neg}L\to (J\vee L)$’ and ‘$\mathrm{\neg}L$’ entail ‘$J$’, we simply need to check whether there is any valuation which makes both ‘$\mathrm{\neg}L\to (J\vee L)$’ and ‘$\mathrm{\neg}L$’ true whilst making ‘$J$’ false. So we use a truth table:
$J$  $L$  $\mathrm{\neg}$  $L$  $\to $  $(J$  $\vee $  $L)$  $\mathrm{\neg}$  $L$  $J$ 
T  T  F  T  T  T  T  T  F  T  T 
T  F  T  F  T  T  T  F  T  F  T 
F  T  F  T  T  F  T  T  F  T  F 
F  F  T  F  F  F  F  F  T  F  F 
The only row on which both‘$\mathrm{\neg}L\to (J\vee L)$’ and ‘$\mathrm{\neg}L$’ are true is the second row, and that is a row on which ‘$J$’ is also true. So ‘$\mathrm{\neg}L\to (J\vee L)$’ and ‘$\mathrm{\neg}L$’ entail ‘$J$’.
We now make an important observation:
If ${\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$1$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$2$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathrm{\dots}$}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$n$}}$ entail $\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{C}$}$ in TFL, then ${\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$1$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$2$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathrm{\dots}$}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$,$}{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{A}$}}_{\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$n$}}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\therefore $}\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\mathcal{C}$}$ is valid.
Here’s why. Suppose ${\mathcal{A}}_{1},{\mathcal{A}}_{2},\mathrm{\dots},{\mathcal{A}}_{n}$ entail $\mathcal{C}$ in TFL. Can it happen that the argument ${\mathcal{A}}_{1},{\mathcal{A}}_{2},\mathrm{\dots},{\mathcal{A}}_{n}\therefore \mathcal{C}$ is invalid? Then there would be a case which makes all of ${\mathcal{A}}_{1},{\mathcal{A}}_{2},\mathrm{\dots},{\mathcal{A}}_{n}$ true and $\mathcal{C}$ false, relative to some symbolization key. This case would generate a valuation of the sentence letters occurring in ${\mathcal{A}}_{1},{\mathcal{A}}_{2},\mathrm{\dots},{\mathcal{A}}_{n}$, and $\mathcal{C}$: take the truth value of any sentence letter to be just the truth value of the corresponding sentence in the case in question. This valuation would also make ${\mathcal{A}}_{1},{\mathcal{A}}_{2},\mathrm{\dots},{\mathcal{A}}_{n}$ true and $\mathcal{C}$ false, since truth values are determined from the truth values of sentence letters by the truth tables of the connectives, and the truth values of the sentence letters are the same in the valuation as they are in the (imagined) case. But this is impossible, since we’ve assumed that ${\mathcal{A}}_{1},{\mathcal{A}}_{2},\mathrm{\dots},{\mathcal{A}}_{n}$ entail $\mathcal{C}$ in TFL, and so there is no valuation which makes all of ${\mathcal{A}}_{1},{\mathcal{A}}_{2},\mathrm{\dots},{\mathcal{A}}_{n}$ true and also makes $\mathcal{C}$ false. So it can’t happen that the argument ${\mathcal{A}}_{1},{\mathcal{A}}_{2},\mathrm{\dots},{\mathcal{A}}_{n}\therefore \mathcal{C}$ is invalid; consequently, it must be valid.
In short, we have a way to test for the validity of English arguments. First, we symbolize them in TFL; then we test for entailment in TFL using truth tables.
12.5 The double turnstile
In what follow, we will use the notion of entailment rather a lot in this book. It will help us, then, to introduce a symbol that abbreviates it. Rather than saying that the TFL sentences ${\mathcal{A}}_{1}$, ${\mathcal{A}}_{2}$, … and ${\mathcal{A}}_{n}$ together entail $\mathcal{C}$, we will abbreviate this by:
The symbol ‘$\models $’ is known as the double turnstile, since it looks like a turnstile with two horizontal beams.
Let’s be clear. ‘$\models $’ is not a symbol of TFL. Rather, it is a symbol of our metalanguage, augmented English (recall the difference between object language and metalanguage from chapter 8). So the metalanguage sentence:
is just an abbreviation for this metalanguage sentence:
The TFL sentences $\mathcal{A}$ and $\mathcal{A}\to \mathcal{B}$ entail $\mathcal{B}$
Note that there is no limit on the number of TFL sentences that can be mentioned before the symbol ‘$\models $’. Indeed, we can even consider the limiting case:
This says that there is no valuation which makes all the sentences mentioned on the left side of ‘$\models $’ true whilst making all sentences on the right side (in this case, $\mathcal{C}$) false. Since no sentences are mentioned on the left side of ‘$\models $’ in this case, this just means that there is no valuation which makes $\mathcal{C}$ false. Otherwise put, it says that every valuation makes $\mathcal{C}$ true. Otherwise put, it says that $\mathcal{C}$ is a tautology. Equally, to say that $\mathcal{A}$ is a contradiction, we can write:
For this says that no valuation makes $\mathcal{A}$ true.
Sometimes, we will want to deny that there is a tautological entailment, and say something of this shape:
it is not the case that ${\mathcal{A}}_{1},\mathrm{\dots},{\mathcal{A}}_{n}\models \mathcal{C}$
In that case, we can just slash the turnstile through, and write:
This means that some valuation makes all of ${\mathcal{A}}_{1},\mathrm{\dots},{\mathcal{A}}_{n}$ true whilst making $\mathcal{C}$ false. Note that it does not follow that ${\mathcal{A}}_{1},\mathrm{\dots},{\mathcal{A}}_{n}\models \mathrm{\neg}\mathcal{C}$, for it is possible that some other valuation makes all of ${\mathcal{A}}_{1},\mathrm{\dots},{\mathcal{A}}_{n}$ true and makes $\mathcal{C}$ true. For instance, $P\u22adQ$ but also $P\u22ad\mathrm{\neg}Q$.
12.6 ‘$\models $’ versus ‘$\to $’
We now want to compare and contrast ‘$\models $’ and ‘$\to $’.
Observe: $\mathcal{A}\models \mathcal{C}$ iff no valuation of the sentence letters makes $\mathcal{A}$ true and $\mathcal{C}$ false.
Observe: $\mathcal{A}\to \mathcal{C}$ is a tautology iff no valuation of the sentence letters makes $\mathcal{A}\to \mathcal{C}$ false. Since a conditional is true except when its antecedent is true and its consequent false, $\mathcal{A}\to \mathcal{C}$ is a tautology iff no valuation makes $\mathcal{A}$ true and $\mathcal{C}$ false.
Combining these two observations, we see that $\mathcal{A}\to \mathcal{C}$ is a tautology iff $\mathcal{A}\models \mathcal{C}$. But there is a really, really important difference between ‘$\models $’ and ‘$\to $’:

‣
‘$\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\to $}$’ is a sentential connective of TFL.

‣
‘$\colorbox[rgb]{0.992156862745098,0.956862745098039,0.980392156862745}{$\models $}$’ is a symbol of augmented English.
Indeed, when ‘$\to $’ is flanked with two TFL sentences, the result is a longer TFL sentence. By contrast, when we use ‘$\models $’, we form a metalinguistic sentence that mentions the surrounding TFL sentences.
The English word ‘implies’ is often used as a synonym for ‘entails’, i.e., ‘$\models $’. Some logicians also use it for the conditional, i.e., ‘$\to $’. Since this is liable to commit the confusion warned about above, we avoid the use of ‘implies’ as much as possible. (Whenever it is used, it means ‘entails’.)
Practice exercises
A. Revisit your answers to exercise 11A. Determine which sentences were tautologies, which were contradictions, and which were neither tautologies nor contradictions.
B. Use truth tables to determine whether these sentences are jointly satisfiable, or jointly unsatisfiable:

1.
$A\to A$, $\mathrm{\neg}A\to \mathrm{\neg}A$, $A\wedge A$, $A\vee A$

2.
$A\vee B$, $A\to C$, $B\to C$

3.
$B\wedge (C\vee A)$, $A\to B$, $\mathrm{\neg}(B\vee C)$

4.
$A\leftrightarrow (B\vee C)$, $C\to \mathrm{\neg}A$, $A\to \mathrm{\neg}B$
C. Use truth tables to determine whether each argument is valid or invalid.

1.
$A\to A\therefore A$

2.
$A\to (A\wedge \mathrm{\neg}A)\therefore \mathrm{\neg}A$

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

4.
$A\vee B,B\vee C,\mathrm{\neg}A\therefore B\wedge C$

5.
$(B\wedge A)\to C,(C\wedge A)\to B\therefore (C\wedge B)\to A$
D. Determine whether each sentence is a tautology, a contradiction, or a contingent sentence, using a complete truth table.

1.
$\mathrm{\neg}B\wedge B$

2.
$\mathrm{\neg}D\vee D$

3.
$(A\wedge B)\vee (B\wedge A)$

4.
$\mathrm{\neg}[A\to (B\to A)]$

5.
$A\leftrightarrow [A\to (B\wedge \mathrm{\neg}B)]$

6.
$[(A\wedge B)\leftrightarrow B]\to (A\to B)$
E. Determine whether each the following sentences are logically equivalent using complete truth tables. If the two sentences really are logically equivalent, write “equivalent.” Otherwise write, “Not equivalent.”

1.
$A$ and $\mathrm{\neg}A$

2.
$A\wedge \mathrm{\neg}A$ and $\mathrm{\neg}B\leftrightarrow B$

3.
$[(A\vee B)\vee C]$ and $[A\vee (B\vee C)]$

4.
$A\vee (B\wedge C)$ and $(A\vee B)\wedge (A\vee C)$

5.
$[A\wedge (A\vee B)]\to B$ and $A\to B$
F. Determine whether each the following sentences are logically equivalent using complete truth tables. If the two sentences really are equivalent, write “equivalent.” Otherwise write, “not equivalent.”

1.
$A\to A$ and $A\leftrightarrow A$

2.
$\mathrm{\neg}(A\to B)$ and $\mathrm{\neg}A\to \mathrm{\neg}B$

3.
$A\vee B$ and $\mathrm{\neg}A\to B$

4.
$(A\to B)\to C$ and $A\to (B\to C)$

5.
$A\leftrightarrow (B\leftrightarrow C)$ and $A\wedge (B\wedge C)$
G. Determine whether each collection of sentences is jointly satisfiable or jointly unsatisfiable using a complete truth table.

1.
$A\wedge \mathrm{\neg}B$, $\mathrm{\neg}(A\to B)$, $B\to A$

2.
$A\vee B$, $A\to \mathrm{\neg}A$, $B\to \mathrm{\neg}B$

3.
$\mathrm{\neg}(\mathrm{\neg}A\vee B)$, $A\to \mathrm{\neg}C$, $A\to (B\to C)$

4.
$A\to B$, $A\wedge \mathrm{\neg}B$

5.
$A\to (B\to C)$, $(A\to B)\to C$, $A\to C$
H. Determine whether each collection of sentences is jointly satisfiable or jointly unsatisfiable, using a complete truth table.

1.
$\mathrm{\neg}B$, $A\to B$, $A$

2.
$\mathrm{\neg}(A\vee B)$, $A\leftrightarrow B$, $B\to A$

3.
$A\vee B$, $\mathrm{\neg}B$, $\mathrm{\neg}B\to \mathrm{\neg}A$

4.
$A\leftrightarrow B$, $\mathrm{\neg}B\vee \mathrm{\neg}A$, $A\to B$

5.
$(A\vee B)\vee C$, $\mathrm{\neg}A\vee \mathrm{\neg}B$, $\mathrm{\neg}C\vee \mathrm{\neg}B$
I. Determine whether each argument is valid or invalid, using a complete truth table.

1.
$A\to B$, $B\therefore A$

2.
$A\leftrightarrow B$, $B\leftrightarrow C\therefore A\leftrightarrow C$

3.
$A\to B$, $A\to C\therefore B\to C$

4.
$A\to B$, $B\to A\therefore A\leftrightarrow B$
J. Determine whether each argument is valid or invalid, using a complete truth table.

1.
$A\vee [A\to (A\leftrightarrow A)]\therefore A$

2.
$A\vee B$, $B\vee C$, $\mathrm{\neg}B\therefore A\wedge C$

3.
$A\to B$, $\mathrm{\neg}A\therefore \mathrm{\neg}B$

4.
$A$, $B\therefore \mathrm{\neg}(A\to \mathrm{\neg}B)$

5.
$\mathrm{\neg}(A\wedge B)$, $A\vee B$, $A\leftrightarrow B\therefore C$
K. Answer each of the questions below and justify your answer.

1.
Suppose that $\mathcal{A}$ and $\mathcal{B}$ are logically equivalent. What can you say about $\mathcal{A}\leftrightarrow \mathcal{B}$?

2.
Suppose that $(\mathcal{A}\wedge \mathcal{B})\to \mathcal{C}$ is neither a tautology nor a contradiction. What can you say about whether $\mathcal{A},\mathcal{B}\therefore \mathcal{C}$ is valid?

3.
Suppose that $\mathcal{A}$, $\mathcal{B}$ and $\mathcal{C}$ are jointly unsatisfiable. What can you say about $(\mathcal{A}\wedge \mathcal{B}\wedge \mathcal{C})$?

4.
Suppose that $\mathcal{A}$ is a contradiction. What can you say about whether $\mathcal{A},\mathcal{B}\models \mathcal{C}$?

5.
Suppose that $\mathcal{C}$ is a tautology. What can you say about whether $\mathcal{A},\mathcal{B}\models \mathcal{C}$?

6.
Suppose that $\mathcal{A}$ and $\mathcal{B}$ are logically equivalent. What can you say about $(\mathcal{A}\vee \mathcal{B})$?

7.
Suppose that $\mathcal{A}$ and $\mathcal{B}$ are not logically equivalent. What can you say about $(\mathcal{A}\vee \mathcal{B})$?
L. Consider the following principle:
Suppose $\mathcal{A}$ and $\mathcal{B}$ are logically equivalent. Suppose an argument contains $\mathcal{A}$ (either as a premise, or as the conclusion). The validity of the argument would be unaffected, if we replaced $\mathcal{A}$ with $\mathcal{B}$.
Is this principle correct? Explain your answer.