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A proof of the deduction theorem for logical formulae in l0 using induction. The theorem states that if a formula β can be derived from a set of hypotheses γ and a formula α, then the implication formula (α → β) can be derived from the hypotheses γ. The document also discusses the importance of the theorem and its implications for logical deductions.
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6.6 The Deduction Theorem for L 0
For any Γ ⊆ Form(L 0
) and
for any α, β ∈ Form(L 0
if Γ ∪ {α} ⊢ β then Γ ⊢ (α → β).
Proof:
We prove by induction on m:
if α 1
,... , α m
is derivable in L 0
from the hypotheses Γ ∪ {α}
then for all i ≤ m
(α → α i
) is derivable in L 0
from the hypotheses Γ.
m=
Either α 1
is an Axiom or α 1
∈ Γ ∪ {α}.
Case 1: α 1
is an Axiom
Then
1 α 1
[Axiom]
2 (α 1
→ (α → α 1
)) [Instance of A1 ]
3 (α → α 1
is a derivation of (α → α 1
) from hypotheses ∅.
Note that if ∆ ⊢ ψ and ∆ ⊆ ∆
′
, then obviously
′
⊢ ψ.
Thus (α → α 1
) is derivable in L 0
from hypothe-
ses Γ.
Case 2: α 1
∈ Γ ∪ {α}
If α 1
∈ Γ then same proof as above works (with
justification on line 1 changed to ‘∈ Γ’).
If α 1
= α, then, by Example 6.4, ⊢ (α → α 1
hence Γ ⊢ (α → α 1
Let β 1
,... , β r be a derivation in L 0
of
(α → α k
) = β r
from Γ
and let γ 1
,... , γ s be a derivation in L 0
of
(α → (α k
→ α m+
)) = γ s
from Γ.
Then
1 β 1
r-1 β r− 1
r (α → α k
r+1 γ 1
r+s-1 γ s− 1
r+s (α → (α k
→ α m+
r+s+1 ((α → (α k
→ α m+
((α → α k
) → (α → α m+
r+s+2 ((α → α k
) → (α → α m+
)) [MP r+s, r+s+1]
r+s+3 (α → α m+
) [MP r, r+s+2]
is a derivation of (α → α m+
) in L 0
from Γ. 2
6.7 Remarks
rule MP.
So any system that includes A1, A2 and
MP satisfies the Deduction Theorem.
verting any derivation showing Γ ∪ {α} ⊢ β
into one showing Γ ⊢ (α → β).
If Γ ⊢ (α → β) then Γ ∪ {α} ⊢ β.
Proof:
. derivation from Γ
r α → β
r+1 α [∈ Γ ∪ {α}]
r+2 β [MP r, r+1]
6.9 Definition
The sequent calculus SQ is the system where
a proof (or derivation) of φ ∈ Form(L 0
) from
Γ ⊆ Form(L 0
) is a finite sequence of sequents,
i.e. of expressions of the form
SQ
ψ
with Γ ⊢ SQ
φ as last sequent.
Sequents may be formed according to the fol-
lowing rules
Ass: if ψ ∈ ∆ then infer ∆ ⊢ SQ
ψ
MP: from ∆ ⊢ SQ
ψ and ∆
′
⊢ SQ
(ψ → χ)
infer ∆ ∪ ∆
′
⊢ SQ
χ
DT: from ∆ ∪ {ψ} ⊢ SQ
χ infer ∆ ⊢ SQ
(ψ → χ)
PC: from ∆ ∪ {¬ψ} ⊢ SQ
χ and
′
∪ {¬ψ} ⊢ SQ
¬χ infer ∆ ∪ ∆
′
⊢ SQ
ψ
‘PC’ stands for proof by contradiction’
Note: no axioms.
6.10 Example of a proof in SQ
1 ¬β ⊢ SQ
¬β [Ass]
2 (¬β → ¬α) ⊢ SQ
(¬β → ¬α) [Ass]
3 (¬β → ¬α), ¬β ⊢ SQ
¬α [MP 1,2]
4 α, ¬β ⊢ SQ
α [Ass]
5 (¬β → ¬α), α ⊢ SQ
β [PC 3,4]
6 (¬β → ¬α) ⊢ SQ
(α → β) [DT 5]
SQ
((¬β → ¬α) → (α → β)) [DT 6]
So ⊢ SQ
We’d better write ‘Γ ⊢ L 0
φ’ for ‘Γ ⊢ φ in L 0
6.11 Theorem
0
and SQ are equivalent: for all Γ, φ
L 0
φ iff Γ ⊢ SQ
φ.
Proof: Exercise
