* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Jesús Mosterín wikipedia, lookup

Novum Organum wikipedia, lookup

Abductive reasoning wikipedia, lookup

History of the function concept wikipedia, lookup

Model theory wikipedia, lookup

Foundations of mathematics wikipedia, lookup

Gödel's incompleteness theorems wikipedia, lookup

History of logic wikipedia, lookup

List of first-order theories wikipedia, lookup

Interpretation (logic) wikipedia, lookup

Propositional formula wikipedia, lookup

First-order logic wikipedia, lookup

Naive set theory wikipedia, lookup

Combinatory logic wikipedia, lookup

Mathematical proof wikipedia, lookup

Laws of Form wikipedia, lookup

Quantum logic wikipedia, lookup

Peano axioms wikipedia, lookup

Mathematical logic wikipedia, lookup

Principia Mathematica wikipedia, lookup

Axiom of reducibility wikipedia, lookup

Intuitionistic logic wikipedia, lookup

Law of thought wikipedia, lookup

Propositional calculus wikipedia, lookup

Transcript

deduction theorem holds for intuitionistic propositional logic∗ CWoo† 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. 1 – 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. References [1] J. W. Robbin, Mathematical Logic, A First Course, Dover Publication (2006) 2