8.5: Reductio Ad Absurdum
Suppose you were given as a premise that statement A implies the
contradiction B • ∼ B, i.e., that A → (B • ∼ B). Using CP and MT it is
easy to show that from this premise ∼ A follows:
1.
2.
3.
4.
5.
6.
7.
A → (B • ~B)
B
~~B
B → ~~B
~B ∨ ~~B
~(B • ~B)
~A
∴ ~A
Assume
2, DN
2-3 CP
4 MI
5 DeM
1,6 MT
In the same way, if we’d assumed that ∼ A implies a contradiction
we could have proved A. The basic insight here is this:
Whatever implies a contradiction must be false!
This principle is embodied in a proof method known as reductio ad
absurdum (RAA), which we now add to our system of propositional
logic. RAA comes in two general forms:
<premises and derived WFFs>
<premises and derived WFFs>
.
.
.
m.
.
.
.
m.
p
Assume
.
.
.
.
.
.
n.
n.
q • ~q
n+1. ~p
m-n RAA
~p
Assume
q • ~q
n+1. p
m-n RAA
Note that when the conclusion of an argument is the negation of a
statement, your assumption should be the unnegated portion of the
statement and try RAA.
Example: A proof for B ↔ ∼ A, ∼ A → ∼ C, C ∨ D, ∼ C → ∼ D ∴ ∼ B
1. B ↔ ∼ A
2. ∼ A → ∼ C
3. C ∨ D
4. ∼ C → ∼ D ∴ ∼ B
Note that it wouldn’t be a mistake to assume ∼∼ B in line 5, but it
would add an unnecessary step, since you would first need to apply
DN to derive B before you could infer ∼ A as in line 8.
2
Example: A proof for (A ∨ B) ∨ (C • D), (A ↔ E) • (B → F),
K ↔ ∼ (E ∨ F), C → B ∴ ∼ K
1. (A ∨ B) ∨ (C • D)
2. (A ↔ E) • (B → F)
3. K ↔ ∼ (E ∨ F)
4. C → B
∴ ∼K
3
When the conclusion of an argument is an atomic statement and
you see no easy way to prove it using other rules, assume the negation of the statement and try RAA.
Example: A proof for ∼ A → (B • C), B →∼ C ∴ A
1. ∼ A → (B • C)
2. B →∼ C
∴ A
More generally:
Tip 12: If the conclusion of an argument (or, more generally, any
statement you are trying to prove in the course of an argument) is
not a conditional statement, and a direct proof of the statement
looks long or difficult, try RAA.
4
We could in principle dispense with CP and just use RAA. To illustrate:
Example: A proof for Z → (∼ Y → X), Z → ∼ Y ∴ Z → X
1. Z → (∼ Y → X)
2. Z → ∼ Y
∴ Z→X
5
It is often possible to combine RAA and CP in useful ways.
Example: A proof for A → (B ∨ C), D → ∼ C ∴ A → ∼ (D • ∼ B)
1. A → (B ∨ C)
2. D → ∼ C
∴ A → ∼ (D • ∼ B)
6