Logic and mathematics seem to be the only domains where self-evidence manages to rise above triviality

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Logic and mathematics seem to be the only domains where self-evidence manages
to rise above triviality; and this it does, in those domains, by a linking of self-evidence
on to self-evidence in the chain reaction known as proof.
W.V.O Quine
Chapter Eight: Derived Rules
INSTRUCTIONAL CONTENT
So far, we’ve learned ten rules, two for each logical operator. Each of these rules is called a ‘basic inference rule,’
and each rule relies upon the properties of the logical operator that the rule features. The rules you’ll learn in this
chapter are called ‘derived rules,’ but, truly, they’re technically basic inference rules as well (just not “in” and
“out” rules.) The nice thing about the rules presented in this chapter is that they provide shortcuts to solving
proofs. In other words, they allow us to solve many of the proofs we worked through before, especially those in
chapters six and seven, in many fewer steps than we could have using the “in” and “out” rules alone. (And none of
them require making a supposition! ☺)
At the end of the notes for this chapter you’ll find a page that includes the form of each derived rule. In fact, I’ve
provided three copies of that page. Here’s why: If you purchased this textbook at the bookstore, I recommend
ripping one of those pages out to use when working through the practice problems at the end of this chapter. This
will save you time in not having to flip back and forth between the practice problem you’re working on and the
page of notes where each rule’s form/structure is first presented. (FYI, the same summary page of derived rules
can be found at the beginning of this textbook, and in the appendix.)
As always, the , , and  are variables:
One important thing to remember when using derived rules (or deciding when and if they can be used) is that
squares, circles, and triangles featured in the forms of these rules (just like in the basic inference rules) are
variables and represent any formula. They can stand for a sentence letter/independent clause such as A, B, C, D…
or they can also stand for longer wffs such as R & M, T → (N & A), (l→P)  (W & V), and so on.
Watch out for dashes:
None of the rules we’ve worked with so far (except for the dash in and dash out rules) included a dash or dashes
in the form of the rule. They only included connectives (→   &) and squares, circles, and triangles (as
variables). The derived rules are slightly different from the basic inference rules in that many of them feature a
dash in the line(s) needed to apply the rule, and/or in the line(s) you’re allowed to derive with that rule.
If a dash appears in the form of a rule, then there must be a dash in the corresponding location of the line you
want to apply the rule to in order to apply that rule. Likewise, if there is a dash (or dashes) that appear in the form
of the rule in the line the rule states you can derive, you must include that dash (or dashes) in the line you derive.
I provide more context for this statement in the Modus Tollens explanation below.
Don’t forget about the basic inference rules
Some of the practice problems at the end of this chapter will be easiest to solve using only derived rules (the rules
presented in this chapter), whereas others are easiest to solve using a combination of derived rules and basic
inference rules (e.g. →O, &I, &O, etc.). I think it’s safe to say, however, that none of the practice problems are
easier to solve by using a supposition-based rule (→I, -I, -O) along with derived rules. Derived rules, in many ways,
allow us to avoid having to use those rules. So don’t make things harder on yourself by using those sorts of rules
along with derived rules. ☺
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The Eight Derived Rules
Modus Tollens (MT)
The MT rule expressed in English: The MT rule expressed symbolically:
If →  appears by itself on an existing line of the proof
and
– is by itself on an existing line of the proof,
then you can derive:
-
x)  → 
y) -
z) - x,y MT
Example:
If Tina needs a BABYSITTER, then she has to WORK today. Tina doesn’t have to work today. Therefore, Tina
doesn’t need a babysitter.
(1) B → W A
(2) -W A
(3) -B 1,2 MT (Modus Tollens (MT) is always justified by two lines.)
Keep in mind that when using Modus Tollens, or any rule for that matter, you cannot rearrange the rule to fit your
needs or alter it in any way. For example, this is an incorrect use of the rule Modus Tollens:
(1) R → A A
(2) -R A
(3) -A 1,2, MT Incorrect!
Although we can’t change the form of a rule (such as is done in the case immediately above, and is the case for
any rule), there is, however, flexibility in terms of what we consider to be represented by, or included in, the 
and the . For example, this is a correct application of Modus Tollens:
(1) J → -V A
(2) – – V A
(3) – J 1,2 MT
In the case above, we can consider J to be inside the  and –V to be inside the . Thus, on our second line of the
proof, – – V is actually the negation of , as –V is represented by the . This allows us to derive –J, or -,
legitimately using Modus Tollens.
Try out these very short proofs, all of which involve the use of MT:
1. Conclusion: -A 3. Conclusion: – – K
1) A → B A 1) – – L A
2) -B A 2) -K → -L A
2. Conclusion: -(E & R) 4. Conclusion: -(J & R)
1) (E & R) → M A 1) (J & R) → (E  G) A
2) -M A 2) -(E  G) A
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Disjunctive Argument (DA)
The DA rule expressed in English: The DA rule expressed symbolically:
If    appears by itself on an existing line of your proof
and
– appears by itself on an existing line of your proof,
then you can derive:

x)   
y) -
z)  x,y DA
Likewise, DA works in this way as well:
If    appears by itself on an existing line of your proof
and
– appears by itself on an existing line of your proof,
then you can derive:

x)   
y) -
z)  x,y DA
Example:
Either WALTER’s wedding is this weekend or BOB’s wedding is this weekend. I just heard that Bob’s wedding
won’t be this weekend. Thus, it must be the case that Walter’s wedding is this weekend.
(1) W  B A
(2) –B A
(3) W 1,2 DA (Disjunctive Argument (DA) is always justified by two lines.)
Based upon our knowledge of the truth properties of a disjunction (which we learned when discussing the
standard truth table and its disjunction column), this rule can be seen as being consistent with the properties of a
disjunction. A disjunction, according to our truth table, is true when each disjunct is true, or when one is true and
one is false (it doesn’t matter which one is true and which is false). The only time a disjunction is entirely false is
when both disjuncts are false. So, consider the rule above as claiming: If you have a disjunction, then it has to be
the case that at least one disjunct is true. If it’s found that one is false (in other words, that - or - is actually
true, thus making  or  false), then the other disjunct must be true and can be derived from those two lines
(the disjunction and the line that includes one negated disjunct).
Try out these very short proofs, all of which involve the use of DA:
1. Conclusion: F 3. Conclusion: – J
1) M  F A 1) -J  M A
2) -M A 2) -M A
2. Conclusion: K 4. Conclusion: E & T
1) K  R A 1) (E & T)  P A
2) -R A 2) -P A
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Conjunctive Argument (CA)
The CA rule expressed in English: The CA rule expressed symbolically:
If -( & ) appears by itself on an existing line of your proof
and
 appears by itself on an existing line of your proof,
then you can derive:
-
x) -( & )
y) 
z) - x,y CA
Likewise, CA works in this way as well:
If -( & ) appears by itself on an existing line of your proof
and
 appears by itself on an existing line of your proof,
then you can derive:
-
x) -( & )
y) 
z) - x,y CA
Example:
It’s not the case that both JANUARY and MOCKTOBER are months of the year. January is a month of the year.
Therefore, Mocktober must not be a month of the year.
Since we’re told in the first premise that January and Mocktober can’t both be months of the year, and we’re also
told in the second premise that January is a month of the year, by deduction (basically) we know that Mocktober
can’t be a month of the year. This is the rationale behind the “conjunctive argument” rule.
(1) –(J & M) A
(2) J A
(3) –M 1,2 CA (Conjunctive Argument (CA) is always justified by two lines.)
This rule can also be seen to work consistently with what we know are the truth properties of conjunctions. The
rule claims that if –( & ) is true and  is true, then - must be true. We can rewrite this as: If it’s not the case
that both  and  are true at the same time but we find that  is true, then we know that  must be false and –
 must be true. It works the same way if in our proof we find  to be true – then we can derive -.
Try out these very short proofs, all of which involve the use of CA:
1. Conclusion: -B 3. Conclusion: – – P
1) -(A & B) A 1) -(E & -P) A
2) A A 2) E A
2. Conclusion: -J 4. Conclusion: -(R & L)
1) -(J & G) A 1) -((R & L) & T) A
2) G A 2) T A
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Chain Rule (CH)
The CH rule expressed in English: The CH rule expressed symbolically:
If  → appears by itself on an existing line of your proof
and
 →  appears by itself on an existing line of your proof,
then you can derive:
 → 
x)  → 
y)  → 
z)  →  x,y CH
Aha! This is the rule we always wanted to use, but weren’t allowed to use, when we learned arrow in. Basically,
this rule allows us to derive a conditional that follows from two conditions that represent a chain – for example, if
we had A → B on one line of our proof and B → C on another line, if we wanted to derive A → C as our conclusion
or as another line of the proof we had to suppose A and then show C follows from it by using arrow out several
times and then we’d use arrow in. The Chain Argument (CH) rule allows us to cut out many obvious steps.
Example:
(1) A → B A Conclusion: A → D
(2) B → C A
(3) C → D A
(4) A → C 1,2 CH (Chain Argument (CH) is always justified by two lines.)
(5) A → D 4,3 CH (Since it is justified by only two lines, this is why we need to apply it
twice.)
The other way:
1 (1) A → B A
2 (2) B → C A
3 (3) C → D A
4 (4) A S
1,4 (5) B 1,4 →O
1,2,4 (6) C 2,5 →O
1,2,3,4 (7) D 3,6 →O
1,2,3 (8) A → D 4,7 →I
Not only can we complete this sort of proof in
fewer steps when using CH, we also avoid
needing a dependency column because we do
not have to suppose A to construct a
conditional between A and D.
Try out these very short proofs, all of which involve the use of CH:
1. Conclusion: A → C 3. Conclusion: (A & B) → L
1) A → -B A 1) (A & B) → C A
2) -B → C A 2) C → L A
2. Conclusion: E → N 4. Conclusion: (M  P) → -R
1) G → N A 1) -T → -R A
2) E → G A 2) (M  P) → -T A
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Double Negation (DN)
The DN rule expressed in English: The DN rule expressed symbolically:
If – –  appears by itself on an existing line of your proof
then you can derive:

x) – – 
y)  x DN
Likewise, DN works in this way as well:
If  appears by itself on an existing line of your proof
then you can derive:
– – 
x) 
y) – –  x DN
This rule allows us to derive a line of the proof that is otherwise identical to a pre-existing line, but has two fewer
or two more dashes modifying the formula that we’ve copied from the former line of the proof. This rule comes in
quite handy when you want to use another rule, but don’t have the exact lines you need to apply that rule (when
deriving a formula with two more or fewer dashes would be the line needed to apply another rule). This is what I
mean:
In the proof below, we can’t apply →O to lines one and two because the formula before the arrow on line one
isn’t identical to the formula on line two. But if we apply the DN rule to line two, we can derive a formula that
then is identical to the antecedent of line one, allowing us then to apply the →O rule:
(1) – – Y → L A
(2) Y A
(3) – – Y 2 DN (Double Negation (DN) is always justified by only one line.)
(4) L 1,3 → O
Remember, any well-formed-formula (wff) can be represented by the , so this is also an appropriate application
of DN:
(1) – – – R A
(2) – R 1 DN
Try out these very short proofs, all of which involve the use of DN:
1. Conclusion: – – M 3. Conclusion: A
1) M A 1) -A → -B A
2) – – B A
2. Conclusion: E & R 4. Conclusion: E
1) – – (E & R) A 1) E  – S A
2) S A
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DeMorgan’s Law (DM)
There are four versions of DeMorgan’s Law (DM):
From -( & ), derive -  - and vice versa
From -(  ), derive - & - and vice versa
From -(- & -), derive    and vice versa
From -(-  -), derive  &  and vice versa
By “vice versa” above, I mean that, for example, the first version of the rule can be applied when you have -( &
) on a line to derive -  -, but the first version of the rule also can be used if you have -  - on a line of
your proof to derive -( & ).
The rules for DeMorgan’s Law are consistent with what was presented in chapter seven regarding the
symbolization of sentences including the phrase “it’s not the case that both” as well as sentences involving the
word “neither.” There is equivalence between various forms of conjunctive formulae and disjunctive formulae,
and this is a rule that captures that logical equivalence. For instance, the first rule of DM states that if we have a
sentence that says that “it’s not the case that BOTH X and Y are the case” then we can derive the logically
equivalent sentence that states “either it’s not the case that X or it’s not the case that Y”. The second, third, and
fourth versions of DM also reflect the logical equivalency between certain conjunctive and disjunctive sentences.
For instance, consider the following sentence:
It’s not the case that both SUSAN is a man and MARK is a man. -(S & M)
The first version of DM above tells us this is logically equivalent to saying…
Either SUSAN is not a man or MARK is not a man. -S  -M
Another Example:
Think of this argument below saying: If my Oregon Trail character was RALPH, then my character got TYPHOID or
DYSENTERY. My character did not get Typhoid and did not get Dysentery. Therefore, my character was not Ralph.
(1) R → (T  D) A
(2) -T & -D A
(3) -(T  D) 2 DM
(4) -R 1,3 MT
(Note: A proof of the argument above would consist of 19 lines if it were to be completed only with basic
inference rules. This way it’s only four.)
When is this rule useful?
Anytime you need to convert a conjunction into a disjunction (or vice versa), take a look at the versions of the DM
rule and see whether one of those versions might be helpful.
Try out these very short proofs, all of which involve the use of DM:
1. Conclusion: -E  -R 3. Conclusion: -(M & E) & -T
1) -(E & R) A 1) -((M & E)  T) A
2. Conclusion: -(-J & -F) 4. Conclusion: L  H
1) J  F A 1) -(-L & -H) A
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Arrow (AR)
There are also four versions of the Arrow (AR) rule:
From  → , derive -   and vice versa
From - → , derive    and vice versa
From  →  derive -( & -) and vice versa
From -( → ) derive  & - and vice versa
By “vice versa” above, I mean that, for example, the first version of the rule can be applied when you have  →
 on a line to derive -  , but the first version of the rule also can be used if you have -   on an existing
line of the proof to derive  → .
The Arrow rule allows us to derive, on a new line of the proof, a formula that is logically equivalent to a preexisting line. In particular, this rule’s versions all express equivalence between a particular conditional and a
particular disjunction or conjunction. For example, consider the following sentence:
If it’s MONDAY, then I have to go to SCHOOL.
The first version of the arrow (AR) rule claims that the sentence above is equivalent to the following:
Either it’s not MONDAY or I have to go to SCHOOL.
In proof form:
(1) M → S A
(2) –M  S 1AR
Notice that in the following proof, even though there isn’t a specific version of the AR rule that tells us we can
apply the rule to a conditional with two dashes modifying the antecedent, we nevertheless can apply the AR rule,
in this case, version two of the AR rule stated above:
(1) – – M → K A
(2) -M  K 1 AR
When is this rule useful?
Anytime you need to convert a conditional into a disjunction or conjunction (or vice versa), take a look at the
versions of the AR rule and see whether one of those versions might be helpful.
Try out these very short proofs, all of which involve the use of AR:
1. Conclusion: -A  B 3. Conclusion: G & -E
1) A →B A 1) -(G → E) A
2. Conclusion: -R → F 4. Conclusion: -A → -C
1) R  F A 1) A  -C A
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Contraposition (CN)
(The last rule! Whew!)
There are three versions of the Contraposition (CN) rule:
From → , derive - → - (and vice versa)
From  → -, derive  → -.
From - → , derive - → .
Every version of the CN rule allows us to derive a logically equivalent conditional from a conditional already
appearing by itself on a previous line of the proof. Also, in every version, the rule requires that in the line we
derive, we flip around the location of the antecedent and consequent of the conditional in comparison to the
previous line of the proof. Lastly, dashes are featured in each version of CN. Be mindful of this, as each application
of CN requires us to add dashes, remove dashes, or leave them in place. Consider the following sentence:
If HANK is at the bank, then RALPH is at the bank. H → R
The first version of the contraposition (CN) rule claims that from the sentence above it follows that…
If RALPH isn’t at the bank, then HANK isn’t at the bank. -R → -H
Which actually makes sense. Let’s say Hank goes to the bank. Then we know Ralph is going to go. But Ralph can
certainly go more than Hank goes – all we know is that every time Hank goes, Ralph follows after him. But what if
Ralph hasn’t gone to the bank at all? Then we know that Hank certainly hasn’t gone, because Ralph hasn’t gone
either on his own time or following Hank, and we know every time Hank goes, Ralph follows.
In proof form:
(1) H → R A
(2) -R → -H 1 CN
When is this rule useful?
Anytime you need to convert a conditional into a very similar conditional but where the antecedent and
consequent are in reverse order, take a look at the versions of the CN rule and see whether one of those versions
might be helpful. Remember, though, to be mindful of what the particular version of the rule you’d like to use
states you need to do with dashes (add, remove, or keep in place).
Try out these very short proofs, all of which involve the use of CN:
1. Conclusion: -B → -A 3. Conclusion: T → -S
1) A → B A 1) – – S → -T A
2. Conclusion: M → -E 4. Conclusion: -G → (A & F)
1) E → -M A 1) -(A& F) → G A