Details of Hunter’s “Informal” Proof of Craig’s Interpolation Theorem for
P
Branden Fitelson
02/06/05
Hunter’s proof of Craig’s Interpolation theorem for
P
is a bit opaque. Here’s a more detailed version of his
proof, which I sketched in class on Friday. Since this is our first (nontrivial) metatheorem, it’s worth doing
a handout that proves it in some detail. We’ll see similar kinds of proofs often in the course.
Theorem
.
Let
A
and
B
be formulas of
P
, such that (1) they share at least one propositional symbol
in common, 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
).
[Intuitively, if
P
A
B
(and
A
and
B
have
some
symbols in common!) it is always possible to reason from
A
to
B
via
a formula
C
that has no propositional symbols not shared by
A
and
B
. This is sometimes called
“linear reasoning” from
A
to
B
, since it takes no “detours” through “irrelevant” or “tangent” unshared
propositional symbols.]
Proof.
Case 1
: There are zero propositional symbols occuring in
A
that do not also occur in
B
. That is, the
set of propositional symbols in
A
is a subset of those in
B
. If we let
S A
be the set of propositional symbols
occurring in a formula
A
, then we can express this case as the case in which
S A
S B
. In this case, just
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 proposi
tional symbols that occur in both
A
and
B
. But, this follows from the fact that
C
A
, and the assumption
of this Case, which is that
S A
S B
. Hence,
S A
S C
S A
S B
, which completes
Case 1
.
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 Spring '07
 FITELSON
 Logic, A1 A2 A1, propositional symbols, A2 A1 A2

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