10.2 Hypothetical Syllogism and Exportation
Two useful rules which we can prove using Conditional Proof are Hypothetical Syllogism (HS) and Exportation (EXPO). We can also present another rule for dealing with disjunctions, Disjunction Elimination.
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10.2.1 Disjunction Elimination Rule
If two roads arrive at the same location, it doesn’t matter which you take.
Two Paths Converge
Suppose I don’t know whether I’m having a boy or a girl. Still, I know I’m having a child. Or, suppose that I don’t know whether I’m going to take the 101 or the 202 to work today. Still, I know that I’m going to work. Or, suppose that I don’t know whether a Republican or Democrat will be elected to office. Still, I know that, either way it goes, somebody will be elected to the office.
Disjunction Elimination is a rule that expresses the idea that, when two paths converge on a single point, then even if you don’t know which path you are on, you know where you will end up. More formally, it says that:
1. P v Q
2. P => R
3. Q => R
C. R (1-3, vE)
That is, if you know that either P or Q, and you know that if P then R, and if Q then R, then you know that R. The rule can also be expressed in terms of a series of assumptions:
1. P v Q
2. | P
3. | R
4. | Q
5. | R
6. R (1, 2-3, 4-5, vE)
Here, we don’t know whether P or Q, but if when we assume P we can deduce R, and then when we assume Q we can deduce R, then we know that, either way, R is true. We cite the initial disjunction first (1), then the lines where we proved R followed from one disjunct (2-3), then the lines where we proved R follows from the other disjunct (4-5). The two ways of expressing vE are equivalent, since, because of the rule of CP, we know that a conditional is the same as proving something follows from a provisional assumption; which form of vE is more useful depends on the premises we have to work with and what we are trying to prove.
We can prove that vE is a valid rule of inference by truth table. There is no row where all of the premises are true yet the conclusion is false:
| P | Q | R | 1. P v Q | 2. P => R | 3. Q => R | C. R |
| T | T | T | T | T | T | T |
| T | T | F | T | F | F | F |
| T | F | T | T | T | T | T |
| T | F | F | T | F | T | F |
| F | T | T | T | T | T | T |
| F | T | F | T | T | F | F |
| F | F | T | F | T | T | T |
| F | F | F | F | T | T | F |
Using Disjunction Elimination
Disjunction Elimination is a rule which can only be used when you have a disjunction and you can prove some common conclusion R no matter which of the two disjuncts is true. Despite the name, it doesn’t say that you can “eliminate” the disjunct by simply concluding one or the other disjunct: instead, you have to be able to show both disjuncts lead to the same conclusion. Here are two examples of proofs which use vE.
A Simple Argument
“Either I’m having a boy or a girl. If I’m having a girl, I’m having a kid; if I’m having a boy, I’m having a kid. Therefore, I’m having a kid.”
Let:
B = I’m having a boy
G = I’m having a girl
K = I’m having a kid
Proof:
1. B v G (basic)
2. G => K (basic)
3. B => K (basic) / C. K
4. K (1-3, vE)
A More Complex Argument
“Either Olivia is defeated or Peter wins the gold Medal. If Peter wins the gold medal, then Peter is happy. It is not both the case that Olivia is defeated and Olivia is happy. Therefore, either Peter is happy or Olivia is not happy.”
Let:
D = Olivia is defeated
G = Peter wins the gold medal
P = Peter is happy
O = Olivia is happy
Proof:
1. D v G (basic)
2. G => P (basic)
3. ~(D & O) (basic) / C. P v ~O
4. | D
5. | ~D v ~O (3 DEM)
6. | ~~D (4 DN)
7. | ~O (5, 6 DS)
8. | P v ~O (7 vI)
9. | G
10. | P (2, 9 MP)
11. | P v ~O (10 vI)
12. P v ~O (1, 4-8, 9-11)
Notice that we have to prove P v ~O twice, first from the provisional assumption of D, then from the provisional assumption of G. We then know that, whether D or G, it follows that P v ~O. So, even if we don’t know whether D or G, if we know that either D or G, then we know that P v ~O.
10.2.2 The Rule of Hypothetical Syllogism
A trail from MO to WY, and from WY to OR, is a trail from MO to OR.
A Chain of Conditionals
The rule of Hypothetical Syllogism (HS) allows us to link up a chain of conditionals. For instance, if we know that if Paul gets smallpox, then Quincy gets smallpox, and we know that if Quincy gets smallpox, then Retha gets smallpox, then we know that If Paul gets smallpox, then Retha gets smallpox.
More formally:
1. P => Q
2. Q => R
C. P => R (1, 2 HS)
Note that the order of the first and second premise doesn’t matter, just like other inferences in propositional logic. So, this is also valid:
1. Q => R
2. P => Q
C. P => R (1, 2 HS)
To make sure you are using HS correctly, be sure to identify the ‘middle term’ which is common between both lines: Q above is the antecedent of one line, and the consequent of the other line. The conclusion then eliminates the middle term; the antecedent (P) remains the antecedent, and the consequent (R) remains the other antecedent.
We will allow multiple applications of HS in a row, provided that each of the individual applications would have been valid. For instance we’ll accept putting two uses of HS in a single step on line 4:
1. A => B
2. B => C
3. C => D
4. A => D (1-3 HS)
How do we know HS is valid? Well, we can prove that HS is valid by using Conditional Proof (CP):
1. P => Q
2. Q => R
3. | P
4. | Q (1, 3 MP)
5. | R (2, 4 MP)
6. P => R (3-5 CP)
Examples of Hypothetical Syllogism (HS)
Here are a few examples of hypothetical syllogism:
Simple Example
“If Shea drinks a lot, then Shea is drunk. If Shea is drunk, then it isn’t safe for Shea to drive a vehicle. Therefore, if Shea drinks a lot, then it isn’t safe for Shea to drive a vehicle”
Let:
S = Shea drinks a lot.
D = Shea is drunk.
V = It is safe for Shea to drive a vehicle.
Proof:
1. S => D
2. D => ~V / C. S => ~V
C. S => ~V (1, 2 HS)
Complex example
“If Shea neither listens to the voice of reason nor his mother, then Shea drinks a lot. If it isn’t safe for Shea to drive a vehicle, then Bob has to drive Shea home. If Shea drinks a lot, then Shea is drunk. If Shea is drunk, then it isn’t safe for Shea to drive a vehicle. Therefore, if Shea neither listens to the voice of reason nor his mother, then it isn’t safe for Shea to drive home.”
Let:
R = Shea listens to the voice of reason.
M = Shea listens to his mother
B = Bob has to drive Shea home.
Proof:
1. ~(R v M) => S
2. ~V => B
3. S => D
4. D => ~V / C. ~(R v M) => B
5. ~(R v M) => D (1, 2 HS)
6. D => B (2, 4 HS)
7. ~(R v M) => B (5, 6 HS)
10.2.3 The Rule of Exportation
If you pass both customs and passport control, then you may enter. So, if you pass customs, then if you pass passport control you may enter.
The Rule of Exportation (EXPO)
The rule of exportation says that a conjunction in the antecedent of a conditional is equivalent to a conditional with one conjunct in the antecedent, and the other conjunct as the antecedent of a conditional in the consequent. In other words, the following two statements are logically equivalent:
((P & Q) => R) ≡ (P => (Q => R))
So, we can infer one from the other:
1. (P & Q) => R
2. P => (Q => R) (1, EXPO)
or
1. P => (Q => R)
2. (P & Q) => R (1, EXPO)
Proving the Rule is Valid
We can prove the rule is valid using conditional proof. We can prove this by first proving that from (P & Q) => R we can validly deduce P => (Q => R), and then proving the other direction, that from P => (Q => R) we can deduce (P & Q) => R, to give us a biconditional. Notice that this means we’ll have to embed one assumption within the scope of another assumption, so there will be two assumption lines (||) instead of one:
1. (P & Q) => R (basic) / C. P => (Q => R)
2. | P (assume)
3. | | Q (assume)
4. | | P & Q (2, 3 &I)
5. | | R (1, 4 MP)
6. | Q => R (3-5 CP)
7. P => (Q => R) (3-6 CP)
Now, let’s prove the other direction:
1. P => (Q => R) (basic) / C. (P & Q) => R
2. | P & Q
3. | P (2 &E)
4. | Q => R (1, 3 MP)
5. | Q (2 &E)
6. | R (4, 5 MP)
7. (P & Q) => R (2-6 CP)
Examples of Exportation
Simple Example
“If you put money in the bank and keep it there, then it will earn interest. So, if you put money in the bank, then if you keep it there, then it will earn interest.”
Let:
P = You put money in the bank
K = You keep money in the bank
E = You earn interest
Proof:
1. (P & K) => E
2. P => (K => E) (1 EXPO)
Complex Example
“If either a bank robbery occurs or a jewelry heist occurs, then, if Malcolm is on duty, then the criminal won’t be caught and the police chief will not be happy. So, if either a bank robbery or a jewelry heist occurs, and Malcolm is on duty, then the criminal won’t be caught and the police chief will not be happy.”
Let:
B = a bank robbery occurs
J = a jewelry heist occurs
M = Malcolm is on duty
C = the criminal will be caught
P = the police chief will be happy
Proof:
1. (B v J) => (M => (~C & ~P))
2. ((B v J) & M) => (~C & ~P) (1 EXPO)
Submodule 10.2 Quiz
Licenses and Attributions
Key Sources:
- Watson, Jeffrey (2019). Introduction to Logic. Licensed under: (CC BY-SA).
Next Page: 10.3 Indirect Proof