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1002
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James Hildebrand
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Philosophy

1002

James Hildebrand

Spring

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SEMANTICS FOR SENTENTIAL LOGIC
Whereas the syntax for SL concerns what strings of vocabulary elements are well formed (are
sentences of SL), the semantics concerns what meaning is to be ascribed to the vocabulary
elements.
These notes are divided into the following sections:
Semantics for atomic sentences of SL
Semantics for compound sentences of SL
Determining the truth values of compound sentences
Determining the truth values of atomic sentences
Definitions of truth-functional properties and relations of sentences and groups of sentences of
SL
Truth functional truth, falsity, and indeterminacy
Truth functional equivalence and non-equivalence
Truth functional inconsistency and consistency
Truth functional entailment and validity
EXERCISES
Comparison of truth functional with logical properties and relations
SEMANTICS FOR ATOMIC SENTENCES OF SL
English sentences, even English simple sentences, have a rich and complex meaning. This is
not the case with sentences of SL. We will say that the meaning of an atomic sentence of SL is
just its truth value. We further specify that there are only two truth values, T and F (or true and
false), and that every atomic sentence of SL must have one or other but not both of these truth
values.
In English, the truth or falsity of a simple sentence is very often determined by the way the worl
d
is. The sentence “Grass is green,” for example, is taken to be true if and only if grass is in fact
green.
In SL the truth or falsity of atomic sentences is determined by truth value assignments. A truth
value assignment assigns either a T or an F, but not both, to every atomic sentence of SL.
Be careful about the difference between a truth value and a truth value assignment. T and F
are truth values. A sentence of SL is assigned one or other of these two truth values by a truth
value assignment. But a truth value assignment, as we are using that expression in this course,
does not just assign one truth value to one atomic sentence of SL. A truth value assignment is
the assignment of one or other of a T or an F to every atomic sentence of SL. The assignment
of one truth value to one atomic sentence of SL is only a part of what a truth value assignment
does.
There are infinitely many truth value assignments. There is the truth value assignment that
assigns a T to every atomic sentence. There is the truth value assignment that assigns a T to
the alphabetically first atomic sentence, “A,” and an F to every other atomic sentence. There is
the truth value assignment that assigns an F to every atomic sentence. There is the truth value
assignment that assigns a T to “A,” “C,” and “H217 and an F to every other atomic sentence.
And so on. In these notes (though not in the text), I will often refer to a truth value assignment by using the
Greek letter “alpha” (α). Where I want to distinguish between different truth value assignments I
will subscript this symbol (α1, α2, 3 , etc.).
We are not concerned which of these infinitely many truth value assignments is the correct one.
There is no correct one. Instead, we will standardly adopt the practice of considering all of
them.
There is an important consequence that must be noted and repeatedly stressed here. The fact
that we are not concerned with which of the infinitely many truth value assignments is the
correct one means that, in SL, truth is relative, not absolute. There is no absolute truth or
falsity. Truth and falsity are relative to truth value assignments.
We will never simply say that a sentence of SL is “true” or “false.” We will only ever say that it is
“true on a truth value assignment,” “false on a truth value assignment,” “true on this truth value
assignment but false on that one,” “true on all truth value assignments,” “true on no truth value
assignment,” and so on. Whenever truth or falsity are ascribed to sentences of SL, it will have
to be relative to one or more truth value assignments, and we will expect that those truth value
assignments will have been mentioned.
Though there are infinitely many truth value assignments, surveying them is not always as big a
job as it might at first seem. If you are just considering one atomic sentence, “A,” then even
though there are infinitely many truth value assignments, they are all of one or other of just two
types, the ones that assign a T to “A” and the ones that assign an F to “A.” We do not need to
worry about what assignments they make to other sentences of SL if we are only worried about
the sentence, “A.” So, in effect, the infinitely many truth value assignments reduce to just two.
Similarly, if you are considering just two atomic sentences, “A” and “B,” then even though there
are infinitely many truth value assignments, they are all of one or other of just four types, the
ones that assign T to both “A” and “B,” the ones that assign T to “A” but F to “B,” the ones that
assign F to “A” but T to “B,” and the ones that assign F to both.
If you are considering just three atomic sentences there are just eight types of truth value
assignments, if just four, sixteen, and so on. In general, the number of different types of truth
value assignments you need to consider is 2 , where n is the number of sentences you are
considering.
These remarks exhaust the semantics of atomic sentences of SL.
SEMANTICS FOR COMPOUND SENTENCES OF SL
The meanings of compound sentences of SL are determined by the meanings of their
component parts. But they are also determined by the meanings of the connectives that are
attached to those component parts.
The meanings of the five connectives of SL are defined in terms of truth values. However, the
connectives do not themselves have truth values. Instead, they determine truth values for the
compound sentences they are used to build, and it is the way they determine these values that constitutes their meaning. The connectives of SL are truth functional connectives. A truth
functional connective creates a compound sentence that has a truth value that is completely
determined by the truth values of its component parts. For each possible combination of truth
values that the component parts might have, the connective must determine exactly one truth
value for the compound. The meaning of the connective is just the rule it uses to determine the
truth value of a compound sentence. Each connective has its own rule, which both tells us what
the connective means and what truth values it assigns to a compound sentence.
~Rule: ~ creates a compound sentence with the opposite truth value of its immediate
component.
&Rule: & creates a compound sentence that is true iff both its immediate components are true
and false iff at least one of its immediate components is false.
vRule: v creates a compound sentence that is true iff at least one of its immediate components
is true and false iff both immediate components are false.
⊃Rule: ⊃ creates a compound sentence that is true iff either its antecedent is false or its
consequent is true and false iff its antecedent is true and its consequent is false.
≡Rule: ≡ creates a compound sentence that is true iff its immediate components have the same
truth value and false iff its immediate components have different truth values.
These rules ensure that every compound sentence of SL must also have one or the other, but
not both of the two truth values, T and F on each possible truth value assignment to its
components. So the meaning of compound sentences ends up being their truth values as well.
The meaning of the connectives is given by the rules I have stated above. It is not given by
what you might imagine are corresponding words of English. You should not think that “&”
means “and” (or “but” or “moreover,” or “however,” etc.). You should not think that “v” means
“or.” And so on. The meaning of the connectives is the rules given above. “&” means “true if
and only if both immediate components are true.” “v” means “true if and only if at least one
immediate component is true.” And so on. There are no words of English that behave, in all
contexts, exactly as specified by these rules. Consequently, if you think of the connectives in
terms of what you might imagine are the corresponding English names, you will only confuse
yourself. Using the formal language, SL, to stand for or “translate” sentences of English or other
natural languages is an art that we will reserve until much later.
DETERMINING THE TRUTH VALUES OF COMPOUND SENTENCES
If we are given a truth value assignment, α, to the atomic sentences of SL, we can use the rules
for the connectives to determine the truth value of the compound ones on α. We do so by
starting with the values α assigns to the atomic components and then, following the same
constructive procedure we would follow were we considering how to construct the sentence
using the syntactic rules, work up from atomic components to the compounds they build. Only
now, we consider what truth value the connective assigns to the compound. Here is an
example of what I mean.
Consider the truth value assignment, α, that assigns T to “A” and “C” and F to “B” and every
other atomic sentence of SL. (We represent this by writing: α(A)=T, α(B)=F, α(C)=T, not
worrying about the other sentences since they aren’t part of the discussion.) Now let’s consider
each of the sentences of SL in 2.1E#2 and calculate their truth values using the rules: 2a.
1. α(B)=F (given)
2. α(A & B)=F (from line 1 by & rule)
2c.
1. α(A)=T (given)
2. α(A v C)=T (from line 1 by v rule)
2e.
1. α(A)=T (given)
2. α(~A)=F (from line 1 by ~ rule)
3. α(~A & ~C)=F (from line 2 by & rule)
2g.
1. α(B)=F (given)
2. α(B & (A v C))=F (from line 1 by & rule)
2i.
1. α(B)=F (given)
2. α((A & C) & B)=F (from line 1 by & rule)
2k.
1. α(C)=T (given)
2. α(B v C)=T (from line 1 by v rule)
3. α((B v C) v ~(B v C))=T (from line 2 by v rule)
Note that in doing these exercises I have adopted the lazy expedient of not doing any more
work than I absolutely have to. The & rule tells me that to prove that a conjunction is false on a
truth value assignment (tva), all I need to do is determine that one of the conjuncts is false on
that tva. So, once I have done that, I do not bother to look into what truth value the other
conjunct has. It will make no difference; the conjunction will have to be false regardless. So I
don’t bother with it and proceed on. You see me do this in 2a, 2e, 2g, and 2i. Indeed, in 2g and
2i you see me get crafty about it and zero in on the one conjunct I know I can show to be false
even though it is not the first one in order or even though its partner is a compound that needs
to be built up before the two can be conjoined.
In 2c and 2k you see me adopt the same lazy method with the v rule. The v rule tells me that t
o
show that a disjunction is true on a tva I only need to prove that one of its disjuncts is true on
that tva. Once I have done that, I can forget about the other disjunct, and that is what I do. In
2c, once I have proven that the first disjunct is true on α, I declare that the whole disjunction is
true on α without bothering to look at the other disjunjct. And in 2k I get crafty about it and zero
in on a path that will allow me to establish the truth value of the whole sentence in the fewest
number of steps. Had I started with “B,” which is false on α, I would not have been in a position
to draw any conclusion about the truth value of the disjunction, because a false disjunct is
compatible with either a true or a false disjunction depending on whether the other disjunct is
true or false. And had I set out to first determine the truth value of “~(B v C)” I would have
wasted even more steps, since that disjunct ends up being false on α and again does not help
me determine the truth value of the whole when the main connective is a “v.” So there was
careful plotting behind my choice of approach. That having been said, all that starting the wrong
place will do is cost you some extra work. Regardless of where you start, as long as you apply
the rules correctly, you will get a result, and it will be the correct one. We can take the same crafty approach to the exercises in 3.1E#3. Here let us consider α to be
the truth value assignment that assigns T to “B” and “C” and F to “A” and every other atomic
sentence of SL.
3a.
1. α(A)=F (given)
2. α(~A)=T (from line 1 by ~ rule)
3. α(~A v (~C v ~B))=T (from line 2 by v rule)
4. α(~[~A v (~C v ~B)])=F (from line 3 by ~ rule)
3c.
1. α(A)=F (given)
2. α(A ⊃ B)=T (from line 1 by ⊃ rule)
3. α((A ⊃ B) v (B ⊃ C))=T (from line 2 by v rule)
3e.
1. α(B)=T (given)
2. α(C)=T (given)
3. α(B ≡ C)=T (from lines 1 and 2 by ≡ rule)
4. α((A ≡ B) v (B ≡ C))=T (from line 3 by v rule)
3g.
1. α(C)=T (given)
2. α(A v C)=T (from line 1 by v rule)
3. α(B ⊃ (A v C))=T (from line 2 by ⊃ rule)
4. α(~[B ⊃ (A v C)])=F (from line 3 by ~ rule)
5. α(~[B ⊃ (A v C)] & ~~B)=F (from line 4 by & rule)
3i.
1. α(A)=F (given)
2. α(B) =T (given)
3. α(~B)=F (from line 2 by ~ rule)
4. α(A ≡ ~B)=T (from lines 1 and 3 by ≡ rule)
5. α(~(A ≡ ~B))=F (from line 4 by ~ rule)
6. α(~A)=T (from line 1 by ~ rule)
7. α(~(A ≡ ~B) ≡ ~A)=F (from lines 5 and 6 by ≡ rule)
8. α(~[~(A ≡ ~B) ≡ ~A])=T (from line 7 by ~ rule)
9. α(B v C)=T (from line 2 by v rule)
10. α(~[~(A ≡ ~B) ≡ ~A] ≡ (B v C))=T (from lines 8 and 9 by ≡ rule)
Note that in 3i there are almost no short cuts that can be taken, so we have no choice but to
look at the truth value α assigns to every atomic component and work up from there. When it
comes to proving that a conjunction is true, a disjunction false, a conditional false, or a
biconditional either true or false there are no shortcuts. You can’t look at just one of the
immediate components to get the answer.
Doing the starred exercises in 2.1E#2 and 3.1E#3 after this same pattern is left to you as an
assignment. DETERMINING THE TRUTH VALUES OF ATOMIC SENTENCES
The rules for the connectives are stated in the form of “if and only if” sentences. This means
that they hold in both directions. Given a truth value assignment to the immediate components
of a sentence, the rules tell you what truth value the compound must have. But equally, given a
truth value assignment to a compound sentence, the rules tell you what truth values its
immediate components must have, and so what truth values the components of those
components must have down to atomic components.
For example, if I am told that “~(A ⊃ B)” is true on a truth value assignment, α, I can reason
“backwards” using the rules as follows:
1. α(~(A ⊃ B))=T (given)
2. α(A ⊃ B)=F (from line 1 by ~ rule)
3. α(A)=T and α(B)=F (from line 2 by ⊃ rule)
Just as the ~ rule tells me that ~P must be true if P is false, so it tells me that ~P is true only if P
is false, from which it follows that just as I can infer the truth of ~P from the falsity of its
immediate component, P, so I can infer the falsity of P from the truth of its negation, ~P.
And just as the ⊃ rule tells me that P ⊃ Q must be false if P is true and Q is false, so it tells me
that P ⊃ Q is false only if P is true and Q is false, from which it follows that just as I can infer the
falsity of P ⊃ Q from the truth of its antecedent, P, and falsity of its consequent, Q, so I can infer
the truth of P and falsity of Q from the fact that a conditional that has P as its antecedent and Q
as its consequent is false.
There is just one wrinkle. When we reason “backwards” from truth values of compounds to truth
values of components like this, we sometimes run into two or even three or more alternatives.
For example, if I am told that “~(A ≡ B)” is true on α, and I reason back, I run into two
alternatives:
1. α(~(A ≡ B))=T (given)
2. α(A ≡ B)=F (from line 1 by ~ rule)
3. α(A)≠α(B) (from line 2 by ≡ rule)
4. Either α(A)=T and α(B)=F, or α(A)=F and α(B)=T (from line 3)
The same follows if I am told that “~(A ≡ B)” is false on α, only now my two alternatives are
either that α assigns a T to both “A” and “B” or that it assigns an F to both of them.
Things are even worse whenever I run into the claim that a conditional or a disjunction is true or
a conjunction is false on a tva. That opens three different possibilities for the way truth values to
the immediate components could go on that tva. This can get messy if the different possibilities
are themselves compound sentences that open further possibilities. All the same, it is worth
looking at how our reasoning would proceed in simpler cases. Let’s consider exercise 2.1#2
again, only this time, let’s assume that each sentence is true on α and attempt to reason back
as far as we can to the truth values of the atomic components. Here is how the unstarred
exercises turn out: 2a.
1. α(A & B)=T (given)
2. α(A)=α(B)=T (from line 1 by & rule)
2c.
1. α(A v C)=T (given)
2. At least one of “A” and “C” is true on α (from line 1 by v rule)
3. Either α(A)=α(C)=T; or α(A)=T and α(B)=F; or α(A)=F and α(B)=T (from line 2)
2e.
1. α(~A & ~C)=T (given)
2. α(~A)=α(~C)=T (from line 1 by & rule)
3. α(A)=α(C)=F (from line 2 by ~rule)
2g.
1. α(B & (A v C))=T (given)
2. α(B)=α(A v C)=T (from line 1 by & rule)
3. α(B)=T and at least one of “A” and “C” is true on α (from line 2 by v rule)
4. Either α (B)=α(A)=α(C)=T; or α (B)=α(A)=T and α (C)=F; or α (B)=α(C)=T and α (A)=F
(from line 3)
2i.
1. α((A & C) & B)=T (given)
2. α((A & C)=α(B)=T (from line 1 by & rule)
3. α(A)=α(C)=α(B)=T (from line 2 by & rule)
2k.
1. α((B v C) v ~(B v C))=T (given)
2. At least one of α(B v C) and α(~(B v C)) is true (from line 1 by v rule)
3. Either α(B v C)=α(~(B v C))=T; or α(B v C)=T and α(~(B v C))=F or α(B v C)=F and
α(~(B v C))=T (from line 2)
4. Either α(B v C)=T and α(B v C))=F; or α(B v C)=T and α(B v C))=T; or α(B v C)=F and
α(B v C)=F (from line 3 by ~ rule)
5. α(B v C)=T or α(B v C)=F (from line 4 because by definition of a truth value
assignment, the first of the three alternatives at line 4 is unacceptable — no
sentence can be both true and false on any truth value assignment — and the
other two include redundancies that have simply been eliminated here)
6. Either at least one of “B” and “C” is true on α or α(B)=α(C)=F (from line 5 by v rule)
7. Either α(B)=α(C)=T; or α(B)=T and α(C)=F; or α(B)=F and α(C)=T; or α(B)=α(C)=F
(from line 6)
Exercise 2k above is interesting for a couple of reasons. You see that what can sometimes
happen as you reason back from a truth value assignment to a compound sentence is not just
that multiple alternatives open up, but that contradictions can emerge that “close” certain
alternatives off. No truth value assignment can assign both a T and an F to the same sentence
(atomic or compound) so when your reasoning leads you to that result, it closes off the
alternative.
This raises an even more interesting possibility. Could it happen that there is no good
alternative? The answer is that it could. To take a simple case, there is no way that “A & ~A”
could be true on any truth value assignment. Any truth value assignment that assigns T to “A” must assign F to “~A” (by ~rule) and so F to “A & ~A” (by & rule). Any truth value assignment
that assigns F to “A” must assign F to “A & ~A.” And all truth value assignments must be of one
or other of these two types. So no tva can make “A & ~A” true.
So what do we say about those cases? If every alternative way of assign
ing a T to a compound
sentence closes off, we know the sentence could not be true on any truth value assignment
whatsoever (not just on α, the particular one we have been considering). If every alternative
way of assigning an F to a compound sentence closes off, we know the sentence could not be
false on any truth value assignment whatsoever. (Note that these are only possibilities when
reasoning back from truth value assignments to compound sentences. By definition of a tva,
any atomic sentence is true on infinitely many tva’s and false on infinitely many others.)
As for the alternative of every way of assigning a T closing off and every way of assigning an F
closing off, that won’t happen. No truth value assignment can leave any atomic sentences
without exactly one truth value, and the connective rules do not permit
any compound sentence
to have anything other than exactly one truth value, given the assignments made to the
components. Consequently, when all the truth values to the atomic components are determined
(as they must be by definition on any tva), all the truth values to the compounds made from
those atoms are determined (as a consequence of the way the rules work), all the compounds
made from those atoms and compounds are determined, and so on up. The possibility of ever
arriving at a compound sentence that does not have exactly one of the two truth values is
foreclosed. This means that if all the ways of assigning a T close off, every possible way of
assigning an F must open up. And when all the ways of assigning an F close off, every possible
way of assigning a T opens up.
This is the other interesting thing about exercise 2k. The sentence, “(B v C) v ~(B v C)” consists
of two atomic components, and we know when there are just two atomic sentences we are
concerned with, there are only 2 =4 possible variants on truth value assignments available to
consider: the ones that make both “B” and “C” true, the ones that make both false, a
nd the ones
that make one or the other true and its partner false. And, as you see, all four are represented
at line 7 of the exercise. What this tells us is that there is no way that this sentence could be
false, because every possible truth value assignment has been found to make it true. And in
fact, if we tried to make it false, we would very quickly find that we run into a contradiction at
every step, illustrating that there can be no truth value assignment that does this.
(The following demonstration adopts the conventions of using “⊥” to designate contradiction, of
indenting lines of demonstration that follow from a supposition, and of “outdenting”
consequences of a supposition — in this case, consequences of having demonstrated that a
supposition leads to an absurdity.)
2k (for false)
1. Suppose: α((B v C) v ~(B v C))=F
2. : α(B v C)=α(~(B v C))=F (from line 1 by v rule)
3. : α(B v C)=F and α(B v C)=T (from line 2 by ~ rule)
4. : ⊥ (from line 3 and the definition of a truth value assignment)
5. α((B v C) v ~(B v C)) cannot =F (from lines 1-4 which demonstrate that the alternative
supposition generates a contradiction)
6. α((B v C) v ~(B v C))=T (from line 5 by definition of a truth value assignment)
Do the starred exercises in 2.1#2 on your own by considering the assigned sentence to be true
on a truth value assignment, α, and then working back to identify all the alternative possible ways α might make assignments to the atomic components. This will help you develop your
familiarity with what has just been said. This will not be a huge task. Keep in mind that none of
the assigned sentences contains more than three atomic components, so at most 8 possibilities
will open up. Usually, there will be considerably fewer.
TRUTH FUNCTIONAL PROPERTIES AND RELATIONS
Truth functional truth, falsity, and indeterminacy
Recall that I said that there are infinitely many truth value assignments. I also said that we do
not care which of these truth value assignments is the correct one. We don’t consider there to
be any one that is more correct than any other.
As we have just seen, many sentences of SL are such that they are true on some truth value
assignments and false on others. We call such sentences truth functionally indeterminate.
To call such sentences “indeterminate” does not mean that their truth value is not determined. It
rather means that it is determined in different ways by different truth value assignments. So,
“indeterminate” is perhaps not the best name for what we mean. But there we go. It is the
name used in the text, so we will follow that practice here.
The atomic sentences of SL are all truth functionally indeterminate. This is because every
possible way of assigning one or other but not both of a T and an F to all the atomic sentences
of SL is a truth value assignment. That means that, for any given atomic sentence of SL, there
is at least one truth value assignment (in fact, there are infinitely many) that makes that
sentence true, and at least one (in fact infinitely many) that makes it false. If nothing else, the
two truth value assignments that make every atomic sentence of SL true and that make every
atomic sentence of SL false will do this.
But, as we have also just seen, while every atomic sentence of SL is true on at least one truth
value assignment and false on at least one other truth value assignment (in fact, infinitely many
of each), there are some compound sentences of SL that are not like this. There are some that
are true on every truth value assignment and others that are false on every truth value
assignment. We will say that a sentence of SL is truth functionally true if and only if it is true
on every truth value assignment. And we will also say that a sentence of SL is truth
functionally false if and only if it is false on every truth value assignment.
The sentence,
A ⊃ A
is truth functionally true. We can prove this as we proved the truth functional truth of sentence
2k above: 1. Suppose: there is a truth value assignment, α, such that α(A ⊃ A)=F
2. : α(A)=T and α(A)=F (from line 1 by ⊃ rule)
3. : ⊥ (from line 2 and the definition of a truth value assignment)
4. It is not the case that there is a truth value assignment, α, such that α(A ⊃ A)=F (from
lines 1-3 which demonstrate that the alternative supposition generates a
contradiction)
5. “A ⊃ A” must be true on every tva, that is, it must be t-f true (from line 4 by definition
of t-f truth).
Note that it would not do to instead suppose that there is a tva, α, such that α(A ⊃ A)=T — at
least not without some careful tweaking.
The reason tweaking would be necessary is that we want to show is that “A ⊃ A” is truth
functionally true, that is, true on every tva. But α is only one tva. So even if we prove that
“A ⊃ A” is true on a tva, α, that is not good enough. T-F indeterminate sentences are also true
on a tva (just not on every tva). The only way to prove that a sentence is T-F true is to prove
that it is true on every tva. Since there are infinitely many truth value assignments, that would
mean either surveying them all, or showing that the opposite, having even one tva that assigns
F, is impossible. That is what we did above.
Admittedly, in the case where a sentence is built up from just one atomic component, all the
infinitely many tva’s reduce to just two: the one’s that assign a T to that component and the
ones that assign an F. This opens the possibility of mounting an argument that “A ⊃ A” must be
truth functionally true because every possible type of truth value assignment assigns a T to it.
Such an argument might run as follows:
1. On any truth value assignment, α, either α(A)=T or α(A)=F (by definition of a tva)
2. Suppose: α(A)=T
3. : then α(A ⊃ A)=T (from line 2 by ⊃ rule — true consequent)
4. Now suppose instead that α(A)=F
5. :then α(A ⊃ A) still =T (from line 2 by ⊃ rule — false antecedent)
6. So, either way, α(A ⊃ A)=T (from lines 1, 2-3 and 4-5)
7. So on any truth value assignment, α, α(A ⊃ A)=T (from lines 1 and 5)
8. “A ⊃ A” must be true on every tva, that is, it must be t-f true (from line 7 by definition
of t-f truth).
While this argument also works, it would have been considerably longer had we been
considering a sentence with two even three (let alone more) atomic components, since then
there would have been four or eight (or exponentially more) different cases to consider. That is
why showing that there is a contradiction in the opposite assumption (in this case, the
assumption that there is a tva that assigns an F) is generally a much more elegant strategy to
adopt. In the remainder of this chapter I will consistently adopt this more elegant strategy,
however “indirect” it may seem. Rather than argue that a sentence is a certain way, I will argue
that if you suppose that it is the opposite way you will run into a contradiction, from which it
follows that it cannot be the opposite way. This argumentative strategy is in fact called “indirect
proof.”
We can use a variant on indirect proof to prove that sentences are T-F false as well as T-F true.
For example, the sentence, A ≡ ~A
is T-F false. Here is an argument that shows this:
1. Suppose: that there is a tva, α, such that α(A ≡ ~A)=T
2. : Then either α(A)=T and α(~A)=T, or α(A)=F and α(~A)=F (from line 1 by ≡
rule)
3. : But then either α(A)=T and α(A)=F, or α(A)=F and α(A)=T (from line 2 by
~ rule)
4. : So, either way, α must assign two different truth values to A (from line 3)
5. : ⊥ (from line 4 and the definition of a truth value assignment)
6. So there can be no tva, α, such that α(A ≡ ~A)=T (from lines 1-5)
7. So“A ≡ ~A” must be t-f false (from line 6 by definition. of t-f falsity)
Proving that a sentence is t-f indeterminate requires an entirely different approach. Rather than
start with a truth value assignment to the whole sentence and reason backwards, we try to think
up two truth value assignments to the atomic parts: one that will make the compound sentence
true and one that will make it false. For example, the sentence,
A ⊃ ~A
is t-f indeterminate. To prove it, I think up two truth value assignments, one that makes it true,
and one that makes it false. I claim that the truth value assignment, α , th1t assigns a T to A,
makes the sentence false, and the truth value assignment, α , that2assigns an F to A makes the
sentence true. Here are my demonstrations:
1. α (1)=T (given)
2. α (1A)=F (1, ~rule)
3. α (1 ⊃ ~A)=F (1, 2, ⊃ rule: true antecedent but false consequent)
4. α (2)=F (given)
5. α (2 ⊃ ~A)=T (4, ⊃ rule: false antecedent)
6. “A ⊃ ~A” has a different truth value on α tha1 it does on α (fro2 lines 3 and 5)
7. “A ⊃ ~A” is t-f indeterminate (from line 6 by def. of t-f indeterminacy)
This was a simple case. In other cases it may be more difficult to think up the truth value
assignments that will do each job. In these cases, what we do is start off by assuming that the
whole sentence is true and reasoning backwards, just as we do when trying to prove a sentence
is t-f false. If the sentence really is t-f indeterminate and not t-f false, then we won’t run into a
contradiction. Instead, we will end up figuring out one or more truth value assignments to the
atomic components that will make the sentence true. Then we can assume the sentence is
false and reason backwards. If the sentence really is t-f indeterminate and not t-f true, we will
find one or more tva’s that will make it false. I illustrate one side of this procedure below, in
connection with proving t-f non-equivalence.
Truth functional equivalence and non-equivalence
We can proceed in the same way to define and determine a truth-functional relation that holds
between sentences of SL, that of truth functional equivalence. We will say that a pair of sentences of SL is truth functionally equivalent if and only if there is
no truth value assignment on which they have different truth values.
A pair of sentences of SL is not truth functionally equivalent if and only if there is at least one
truth value assignment on which they have different truth values.
There is a common misunderstanding of the notion of t-f equivalence that needs to be cleared
up right at the outset. When we say that there is no truth value assignment on which two
sentences of SL, P and Q, receive different truth values, we just mean that there is no tva on
which P and Q have different truth values from one another. We do not mean that there is no
tva on which P and Q have different truth values from any other tva. We do not require that P
must individually receive the same truth value on every truth value assignment (that it must be
either t-f true or t-f false) and that Q must receive this same value on every tva. All that we
require is that, on any truth value assignment you happen to pick, P and Q are both the same
on that assignment. They might both be true on one, and both false on another. As long as
they are both the same on each, the definition has been satisfied. Put in a picture, this suffices
for t-f equivalence,
P Q
α F F
1
α2 F F
α3 T T
α4 F F
α5 T T
α6 F F
• • •
• • •
• • •
supposing that on each of the infinitely many lines that follow, there are either two T’s or two F’s
— supposing there is no line with a T and an F or an F and a T on it.
We do not insist on this:
P Q P Q
α T T α F F
1 1
α 2 T T α 2 F F
α 3 T T α 3 F F
α 4 T T otrhis: α 4 F F
α 5 T T α 5 F F
α 6 T T α 6 F F
• • • • • •
• • • • • •
• • • • • •
As long as on each row they are both the same, its is good. They do not have to be the same
on one row that they are on another.
To prove that a pair of sentences is t-f equivalent there are two alternatives that must be ruled
out. We first suppose that the first sentence is true and the second one is false and show that
this leads to a contradiction. Then we suppose that the second one is false and the first one is true and show that this leads to a contradiction. From this it follows that there is no way the two
sentences can have different truth values.
For example, the pair of sentences,
A ⊃ B ~B ⊃ ~A
are t-f equivalent. Here is the proof:
1. Suppose: there is a tva, α, s.t. α(A ⊃ B)=T and α(~B ⊃ ~A)=F
2. : α(~B)=T and α(~A)=F (1, ⊃ rule applied to false ~B ⊃ ~A)
3. : α(B)=F and α(A)=T (2, ~ rule)
4. : α(A ⊃ B)=F (3, ⊃ rule)
5. : ⊥ (from 1 clause of line 1 and line 4 by definition of a tva)
6. So there cannot be a tva, α, s.t. α(A ⊃ B)=T and α(~B ⊃ ~A)=F (1-5)
7. Suppose: there is a tva, α, s.t. α(A ⊃ B)=F and α(~B ⊃ ~A)=T
8. : α(A)=T and α(B)=F (7 ⊃ rule applied to false A ⊃ B)
9. : α(~A)=F and α(~B)=T (8, ~ rule)
10. : α(~B ⊃ ~A)=F (9, ⊃ rule)
11. : ⊥ (2nd clause of 7, 4, def. of tva)
12. So there cannot be a tva, α, s.t. α(A ⊃ B)=F and α(~B ⊃ ~A)=T (7-11)
13. So no tva can assign different truth values to “A ⊃ B” and “~B ⊃ ~A” (from 6 and 12)
14. So they are t-f equivalent (13, def. of t-f equivalence).
If a pair is not t-f equivalent, one or other or both of these attempts will fail to lead to a
contradiction, and we will end up discovering a truth value assignment on which the two
sentences have different truth values. For example, the sentences,
~(A v B) ~A v ~B
are not t-f equivalent. If we try to prove that they are equivalent, we won’t find the contradictions
we are looking for. In this case, if we suppose that the first sentence is true and the second
sentence is false, we will run into a contradiction.
1. Suppose: there is a tva, α, s.t. α(~(A v B))=T and α(~A v ~B)=F
2. : α(~A)=F and α(~B)=F (1, v rule applied to false ~A v ~B)
3. : α(A)=T and α(B)=T (2, ~ rule)
4. : α(A v B)=T (3, v rule)
5. : α(~(A v B)=F (4, ~rule)
6. : ⊥ (from 1 clause of line 1 and line 4 by definition of a tva)
But, interestingly, we don’t find a contradiction if we suppose that the first sentence is false and
the second sentence is true.
1. Suppose: there is a tva, α, sst. α(~(A v B))=F and α(~A v ~B)=T
2. : α(A v B)=T (1 clause of line 1, ~ rule)
3. : α assigns a T to at least one of “A” and “B” (2, v rule)
4. Suppose: α(A)=T but α(B)=F
5. : α(~B)=T (4, ~ rule)
6. : α(~A v ~B)=T (5, v rule) This argument is incomplete. But as far as it goes, it has uncovered an assignment that α could
make to all the atomic components of the two sentences that makes one false and the other
true without running into the contradiction we were looking for. This suggests that any truth
value assignment that assigns T to “A” and F “B” will also assign different truth values to
“~(A v B)” and “~A v ~B.” Since there is at least one truth value assignment on which these
sentences receive different truth values, and they are not t-f equivalent.
To rigorously prove this, I take the truth value assignment my earlier, failed attempt at proving
equivalence led me to discover, and show by appeal to the semantic rules that it do
es indeed
lead, without producing any contradictions, to an assignment of F to “~(A v B)” and T to
“~A v ~B.” Here is that demonstration:
1. α(A)=T (given)
2. α(A v B)=T (1, v rule)
3. α(~(A v B))=F (3, ~ rule)
4. α(B)=F (given)
5. α(~B)=T (4, ~rule)
6. α(~A v ~B)=T (5, v rule)
7. While “~(A v B)” is false on α “~A v ~B” is true on α (from lines 3 and 6)
8. “~(A v B)” and “~A v ~B” are not t-f equivalent (from line 7 by def. of t-f non-
equivalence)
Truth functional inconsistency and consistency
A set of sentences of SL is truth functionally inconsistent if and only if there is no truth value
assignment that assigns a T to every sentence in that set.
A set of sentences of SL is truth functionally consistent if and only if there is at least one truth
value assignment that assigns a T to every sentence in that set.
Once again, there are common misunderstandings that need to be forestalled at the outset.
For a set to be t-f inconsistent we only require that there be no tva that assigns a T to every
sentence in the set. We do not require that there be a tva that assigns an F to each set
member. That is a far stronger demand than what we are looking for. We only require that
each tva assign an F to at least one set member. Moreover, we do not require that it be the
same set member for each tva that gets the F. A set does not have to contain a t-f false
sentence in order to be t-f inconsistent (though that certainly suffices). It just has to be the case
that one or other of the set members is false on each tva.
In determining truth-functional inconsistency or consistency we are concerned with one and only
one thing: the presence or absence of a truth value assignment that assigns a T to every
sentence in the set. Other truth value assignments are of absolutely no interest to us. In
particular, truth value assignments that assign an F to every sentence in the set are of
absolutely no interest to us. So, given the set, {P, Q, R} of sentences of SL, and the following
truth value assignments, α -1 t7 set the members, P Q R
α 1 F F F
α T F T
2
α 3 F T T
α 4 F T F
α T T T
5
α 6 T F F
α 7 T T F
• • • •
• • • •
• • • •
we declare the set to be t-f consistent. The fact that there is a truth value assign1ent, α , that
makes all the sentences in the set false is absolutely irrelevant to our decision.

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