A Proper Inductive Proof of the Interpolation Theorem for
P
Branden Fitelson
02/14/07
Theorem
. Let
A
and
B
be formulas of
P
, such that (1) they share at least one propositional symbol in com
mon, and (2)
±
P
A
⊃
B
. For any two such formulas of
P
, there exists a formula
C
(called the
P
interpolant
of the formulas
A
and
B
) such that (3)
±
P
A
⊃
C
, (4)
±
P
C
⊃
B
, and (5)
C
contains only propositional
symbols that occur in both
A
and
B
(
i.e.
, only propositional symbols shared by
A
and
B
).
Setup for an inductive proof
. Let
S(A)
be the set of propositional symbols occurring in
A
,
S(B)
be the set
of propositional symbols occurring in
B
, and
q
be some propositional symbol that is shared by
A
and
B
(
i.e.
,
q
∈
S(A)
∩
S(B)
). We will focus on the set
X
=
S(A)

S(B)
of propositional symbols that occur in
A
but not in
B
. We will prove the interpolation theorem by strong mathematical induction on the cardinality
of
X
. That is, we will prove that the following property of natural numbers holds for all
n
≥
0:
S
(n)
: The interpolation theorem (above) holds when
X
=
S(A)

S(B)
=
n
.
Proof.
As always, a proper inductive proof involves a Basis Step and an Inductive step.
Basis Step
. Prove
S
(
0
)
. That is, we must prove the interpolation theorem for the case in which there
are
zero
propositional symbols occurring in
A
that do not also occur in
B
(
X
=
0). In this case, the set
of propositional symbols in
A
is a subset of those in
B
[
S(A)
⊆
S(B)
]. Let
C
=
A
. Then, obviously, (3)
±
P
A
⊃
C
, since
±
P
A
⊃
A
. And, since the assumption of the theorem is that
±
P
A
⊃
B
, we also know that
±
P
C
⊃
B
. All we need to show is that (5)
C
contains only propositional symbols that occur in both
A
and
B
.
But, this follows from the fact that
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 Spring '07
 FITELSON
 Logic, Inductive Reasoning, propositional symbols

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