COMMON COMBINATIONS
There are many useful combinations of logical
operations. Some are so handy and are used so often that they've
acquired standard names. (We needn't commit the names to permanent
memory, but it's very much to our advantage to understand the concepts.)
Exclusive Disjunction (XOR) As we noted when logical operations were being
introduced, the logical OR operation is always considered in the
inclusive sense of the word—i.e., "either one option or the other, or
both." The exclusive sense of OR, "either one or the other,
but not both," has no symbol of its own, despite that it's fairly
common in non-symbolic reasoning. However, it can be expressed
with various combinations of the five available operators. Perhaps
among the first that spring to mind is (p * ~q) ˅ (q * ~p),
"either p and not q, or else q and not p."
The possibility of an "or both" option in this instance is ruled out by
the fact that the two disjuncts, (p * ~q) and (q * ~p),
are mutually contradictory. However, there's another version
that's both more concise and less subject to faulty manipulation: p
↔ ~q.
Recall from our truth-table analysis that a
bi-condition evaluates as true whenever the truth values of both of its
component statements are the same, either both true or both false.
Thus, p ↔
~q evaluates as true if both p and ~q are true
(i.e., p is true and q is false), or if both p and
~q are false (i.e., p is false and q is true), and
evaluates as false whenever p and q are either both true
or both false. Moreover, the behavior of a bi-condition with a
negation is such that it functions exactly the same whether we express it...
p
↔
~q
...or... ~p ↔ q
...or... ~(p ↔ q).
This makes the bi-condition with negation a versatile, concise, and
functionally exact expression of "either p or q, but not
both."
Modus Ponens Consider the IF...THEN condition: p → q.
By itself, it doesn't tell us anything about the actual truth values of
p or q. But suppose we add an assumption that p
is in fact true, expressed simply as: p. Now this tells us
something useful about p → q,
specifically this: IF p is true, THEN q must be true; AND
indeed p is true. Since p is actually true, q
must be true as well! We can express this whole logical relationship
symbolically as a group of three statements:
|
1. |
|
Assumption: |
|
p → q |
|
(if p is true, then q must be
true) |
|
2. |
|
Assumption: |
|
p |
|
(p is true) |
|
3. |
|
Conclusion, from 1 & 2: |
|
q |
|
(so, q is true) |
We can also combine statements, taking one pair
at a time and grouping the operations as we go. Combine the line-1
condition p → q
with the line-2 assumption p, setting the condition off with
parentheses and using conjunction to indicate that both statements are
assumed true:
|
1. |
|
Assumption: |
|
(p → q)
* p |
|
(if p is true, then q must be
true; and p is true) |
|
2. |
|
Conclusion, from 1: |
|
q |
|
(so, q is true) |
If we like, we can also graft the conclusion
onto the statement of assumptions. When we do this, the whole
"if...and...so"
construction takes the form of another condition, with the existing
assumptions bundled together as the antecedent, and the conclusion as
the consequent:
|
((p → q)
* p) →
q |
|
(if it's true, both that if p is
true then q must be true, and also that p is indeed true,
then q is true) |
Symbolically, it makes perfect sense, and (with
a little practice) can be comprehended in little more than the blink of
an eye (versus maybe a dozen blinks for a novice). But the
plain-language interpretation is unwieldy, given the imprecision of language in arranging multiple thoughts into clearly defined yet
interconnected groups. That's
why logicians have cooked up this method of precisely expressing and analyzing
reasoning in the abstract. Not only is it easier on both the eyes
and the brain, but it also does away with writer's cramp! But we
digress...
The matter of true significance from this little
exercise is that here we have a very useful and widely used logical
combination of a condition and the
confirmation of its antecedent...
(p →
q) * p
...to derive the truth of its
consequent...
\q
Or to illustrate the abstraction with a vaguely
familiar concrete example:
"If Paisley goes to prison, then Quigley can enjoy quiet; and Paisley goes
to prison. So, Quigley can enjoy quiet."
Or: "If Grandma picks apples, then she'll bake
an apple pie; and Grandma is picking apples. So, she'll bake an
apple pie."
Or substitute any similarly related statements
for p and q, and the results should be consistent.
Modus Tollens Now recall that, in a conditional relationship, we can also use a false
consequent to draw an inference about an antecedent with an unknown
truth value. Specifically, for any condition where if p is true
then q must be true, it follows (by a logical inference known as
contraposition) that if q is false, then p
cannot be true—and therefore p must be
false. Or, to put it symbolically, p
→ q (IF
p THEN q) is logically equivalent to ~q
→ ~p (IF
NOT q, THEN NOT p). So we can combine the
conditional relationship with other statements thus...
|
1. |
|
Assumption: |
|
p → q |
|
(if p is true, then q must be
true) |
|
2. |
|
Contraposition, from 1: |
|
~q → ~p |
|
(if q is false, then p must
be false) |
|
3. |
|
Assumption: |
|
~q |
|
(q is false) |
|
4. |
|
Conclusion, from 2 & 3: |
|
~p |
|
(so, p is false) |
Again, we can combine the assumptions into a
single statement (and we can omit the contraposition step, since, as we
noted in the IF...THEN truth-table exercise, it's an inherent
implication of a condition, included here only for clarification):
|
1. |
|
Assumption: |
|
(p → q) *
~q |
|
(if p is true, then q must be
true; and q is false) |
|
2. |
|
Conclusion, from 1: |
|
~p |
|
(so, p is false) |
This gives us modus tollens, the opposite
side of the modus ponens relationship.
For both modus ponens and modus
tollens, in each case we use two logical statements taken together
to derive a conclusion. We do this by integrating the logical implications of each statement into an overall conclusion not
implied by either of the individual statements taken alone. The
condition states only a hypothetical relationship, making no claim about
the actual truth values of its component statements. The other
assumption establishes either that the condition's antecedent is true
(modus ponens) or
that its consequent is false (modus tollens). Linking these two pieces of
information by conjunction, we deduce the truth value of the condition's
other statement.
CAUTION! These are the only two valid
implications of an IF...THEN condition. Don't make the all too
common mistake of getting them backward! As truth tables of the
basic operations clearly demonstrated, it's logically impossible to
infer the truth value of a consequent from a false antecedent, or the
truth value of an antecedent from a true consequent. Attempting to
do so results in fallacies of the "formal error" category (denying
the antecedent and affirming the consequent, respectively),
which are responsible for turning countless logical lines of reasoning
into illogical nonsense.
Hypothetical Syllogism We
can use a chain of related IF...THEN operations to determine the
relationship between the first and last links of the chain.
|
1. |
|
Assumption: |
|
p → q |
|
(if p is true, then q must be
true) |
|
2. |
|
Assumption: |
|
q →
r |
|
(if q is true, then r must be
true) |
|
3. |
|
Conclusion, from 1 & 2: |
|
p →
r |
|
(so, if p is true, then r
must be true) |
If we know that whenever one thing is true a
second must be true, and that whenever the second thing is true a third
must also be true, then we can logically conclude that whenever the
first thing is true the third will also be true. This doesn't
nullify the intermediate link in the chain, but if the linkage is well
established and is understood to work every time without fail, we can skip enumerating intermediate steps for convenience's
sake. In some situations, hypothetical syllogism can bridge
several intermediate steps, provided we're well convinced that each step
must indeed always follow from the ones before it.
Consider this multi-step example. A
college freshman might speculate thus about his prospects: "If I earn a
liberal arts degree, then chances are I'll become broadly knowledgeable,
well respected, and sought after. If I become broadly
knowledgeable, well respected, and sought after, then I'll have my
choice of lucrative and satisfying occupations. If I have a
lucrative and satisfying occupation, then I should be able to live
happily." So, assuming the freshman's assumptions are realistic
and no unexpected calamity overwhelms his plans, he's justified in
abbreviating his line of reasoning to, "If I earn a liberal arts degree,
then chances are I should be able to live happily."
Disjunctive Syllogism The
falseness of one disjunct of a disjunction implies the truth of the
other disjunct.
|
1. |
|
Assumption: |
|
p ˅ q |
|
(either p is true or q is
true) |
|
2. |
|
Assumption: |
|
~p |
|
(p is false) |
|
3. |
|
Conclusion, from 1 & 2: |
|
q |
|
(so, q must be true) |
The general claim of a disjunction is that at
least one of its disjuncts must be true. So, if one of the
disjuncts is false, then the other disjunct must be true. If, as
in this example, p is false, then q must be true. Or
if we know that q is false, then p must be true. If
it should ever happen that both disjuncts are false, then the
disjunction itself is exposed as a false claim.
Here's a familiar (if less than air-tight)
example: an ultimatum handed down from each generation to the next.
"You must either go to school or find a job." If you don't go to
school, then you must find a job; if you don't find a job, then you must
go to school. Doing both is also an acceptable option, but doing
neither is not.
Negation of Operations
The negation of an individual statement is
easily enough understood; however, the negation of an operation or a
combination of operations bears a bit of clarification.
When we negate an operation, the conjunction
p * q
for example, we first enclose the operation in parentheses,
(p * q),
and then insert the negation sign before the first parenthesis:
~(p * q).
Now exactly what does this signify? That
p
is false and q
is false? No, not necessarily. The negation of an operation
does not imply the negation of every statement in the operation, but
just the negation of the operational relationship as a whole.
Recall that the implication of the un-negated operation
p * q
is that p
is true and q
is true. Any false truth value for either
p or
q renders the p * q
conjunction operation as a whole false. So
~(p * q),
the negation of p * q,
could have any of three implications: (1) p
is false and q
is false (~p * ~q);
(2) p
is true and q
is false (p * ~q);
or (3) p
is false and q
is true (~p * q).
The expression ~(p * q)
does not tell us which of these conditions is actually the case, but
only that it's not the case that both
p and
q
are true. Thus, while ~p
* ~q is one possible implication of
~(p * q),
it's not the only possibility; so we can't logically jump to that
conclusion.
The situation with a conditional relationship is
similar; ~(p → q)
might mean ~p → ~q,
but it could just as well mean ~p → q
or p → ~q. The only
thing it signifies for certain is that the relationship IF
p THEN
q is not
true, so we'd be unwise to draw any other conclusion without further
information.
With a bi-conditional relationship, the
situation gets a shade more predictable. Recall that a
bi-condition has the interesting characteristic of evaluating as true
whenever both of its conditions have the same truth value—either
both true or both false. Or to put it symbolically,
p ↔ q
implies either p * q
or ~p * ~q.
So, as we might suppose, the negation of a bi-condition implies that its
two component statements have opposite truth values—one true and
the other false; that is, ~(p ↔ q)
implies either ~p * q
or p * ~q.
We've saved the negation of a disjunction for
last, because it's the only sure bet in the lot. We express the
negation of the disjunction p
˅ q using the same procedure for
negating other operations: ~(p
˅ q). Recall that the condition
for a disjunction to evaluate as true is for at least one of its
disjuncts to be true. Another way of expressing this is that the
only condition under which a disjunction evaluates as false is if both
of its disjuncts are false. Thus,
~(p ˅ q)
implies both that p
is false and that q
is false, or simply ~p *
~q, with no other alternatives.
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