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deduction theorem holds for
intuitionistic propositional logic∗
2013-03-22 3:53:41
In this entry, we show that the deduction theorem below holds for intuitionistic propositional logic. We use the axiom system provided in this entry.
Theorem 1. If ∆, A `i B, where ∆ is a set of wff ’s of the intuitionistic propositional logic, then ∆ `i A → B.
The proof is very similar to that of the classical propositional logic, given
here, in that it uses induction on the length of the deduction of B. In fact,
the proof is simpler as only two axiom schemas are used: A → (B → A) and
(A → B) → ((A → (B → C)) → (A → C)).
Proof. There are two main cases to consider:
• If B is an axiom or in ∆ ∪ {A}, then
B, B → (A → B), A → B
is a deduction of A → B from ∆, where A → B is obtained by modus
ponens applied to B and the axiom B → (A → B). So ∆ `i A → B.
• Now, suppose that
A1 , . . . , A n
is a deduction of B from ∆ ∪ {A}, with B obtained from earlier formulas
by modus ponens.
We use induction on the length n of deduction of B. Note that n ≥ 3. If
n = 3, then C and C → B are either axioms or in ∆ ∪ {A}.
– If C is A, then C → B is either an axiom or in ∆. So ∆ `i A → B.
∗ hDeductionTheoremHoldsForIntuitionisticPropositionalLogici created: h2013-03-2i by:
hCWooi version: h42502i Privacy setting: h1i hTheoremi h03F55i h03B20i
† This text is available under the Creative Commons Attribution/Share-Alike License 3.0.
You can reuse this document or portions thereof only if you do so under terms that are
compatible with the CC-BY-SA license.
– If C → B is A, then C is either an axiom or in ∆. Then
E0 , C → (A → C), C, A → C, (A → C) → ((A → (C → B)) → (A → B)),
(A → (C → B)) → (A → B), A → B
is a deduction of A → B from ∆, where E0 is a deduction of the
theorem A → A, followed by an axiom instance, then C, then the
result of modus ponens, then an axiom instance, and finally two
applications of modus ponens. Note the second to the last formula
is just (A → A) → (A → B).
– If C and C → B are axioms or in ∆, then ∆ `i A → B based on the
deduction C, C → B, B, B → (A → B), A → B.
Next, assume there is a deduction E of B of length n > 3. So one of
the earlier formulas is Ak → B, and a subsequence of E is a deduction
of Ak → B, which has length less than n, and therefore by induction,
∆ `i A → (Ak → B). Likewise, a subsequence of E is a deduction of Ak ,
so by induction, ∆ `i A → Ak . With the axiom instance (A → Ak ) →
((A → (Ak → B)) → (A → B)), and two applications of modus ponens,
we get ∆ `i A → B as required.
In both cases, ∆ `i A → B, and the proof is complete.
Remark The deduction theorem can be used to prove the deduction theorem
for the first and second order intuitionistic predicate logic.
[1] J. W. Robbin, Mathematical Logic, A First Course, Dover Publication