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My Rendition of Hunter’s Strong Inductive Proof of The Deduction Theorem for PS
Branden Fitelson
02/13/07
Theorem
. Let
Γ
be an arbitrary set of formulas of P, and let
A
and
B
be arbitrary formulas of P.
If
Γ
∪{
A
} ‘
PS
B
,
then
Γ
‘
PS
A
⊃
B
.
If there is a derivation (in PS) of
B
from
Γ
∪{
A
}
, then there is a derivation (in PS) of
A
⊃
B
from
Γ
alone.
Proof.
We will prove this by
strong induction
on the length of derivations establishing
Γ
∪{
A
} ‘
PS
B
. More
precisely, we will prove by strong induction that the statement
S(n)
is true for all
n
≥
1, where
S(n)
is:
S(n)
:
If
there is a derivation (in PS) of
B
from
Γ
∪{
A
}
of length
n
,
then
there is a derivation (in PS) of
A
⊃
B
from
Γ
alone.
To prove that
S(n)
is true for all
n
≥
1 by strong induction, we will proceed in two steps:
I.
Basis Step
: Prove that
S(
1
)
is true.
II.
Inductive Step
:
Assume
as our
inductive hypothesis
that
S(i)
is true for all
i
such that 1
< i < n
.
Then,
use
this
inductive hypothesis
to show that
S(i)
is true when
i
=
n
(
i.e.
, that
S(n)
is true).
Having accomplished both (I) and (II), we will have succeeded in showing that
S(n)
is true for all
n
≥
1.
In what follows, I will use
D
to stand for some derivation (in PS) of
B
from
Γ
∪ {
A
}
, and I will use
D
0
to
stand for some derivation (in PS) of
A
⊃
B
from
Γ
alone. Since we are concerned about the length of
D
, I
will use
D
1
,
D
2
,...,
D
m
to stand for the
m
lines of a derivation
D
with length
m
. Now, the inductive proof.
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 Spring '07
 FITELSON

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