Download e-book for iPad: Logic: a Brief Course by Daniele Mundici

By Daniele Mundici

ISBN-10: 8847023602

ISBN-13: 9788847023604

This brief publication, geared in the direction of undergraduate scholars of computing device technology and arithmetic, is particularly designed for a primary direction in mathematical good judgment. an evidence of Gödel's completeness theorem and its major results is given utilizing Robinson's completeness theorem and Gödel's compactness theorem for propositional common sense. The reader will familiarize himself with many uncomplicated principles and artifacts of mathematical good judgment: a non-ambiguous syntax, logical equivalence and end result relation, the Davis-Putnam method, Tarski semantics, Herbrand versions, the axioms of id, Skolem general varieties, nonstandard types and, apparently sufficient, proofs and refutations seen as photo gadgets. The mathematical necessities are minimum: the publication is obtainable to anyone having a few familiarity with proofs by means of induction. Many workouts at the courting among common language and formal proofs make the publication additionally attention-grabbing to quite a lot of scholars of philosophy and linguistics.

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We will show that St−1 is satisfied by an appropriate extension αt−1 of αt . To begin, let ω be the extension of αt that assigns value 0 to all variables released in step t. Let ω − and ω + be the two possible extensions of ω on the pivot Pt : in other words, ω + assigns 1 to the pivot, while ω − assigns to it 0. Both ω + and ω − are suitable for St−1 . Claim. At least one among ω + and ω − satisfies St−1 . Suppose by contradiction that the claim does not hold. Then there exist in St−1 two clauses C1 and C2 such that ω − |= C1 and ω + |= C2 .

Putting α1 ⊇ α2 with α1 (X12 ) = 0 it follows that α1 |= S1 . In Step 1 only the pivot is deleted. Putting α0 ⊇ α1 with α0 (X11 ) = 1 we conclude that α0 |= S0 . Following this procedure we have constructed an assignment that satisfies S. We note that S represents the problem of bicolouring three vertices 1, 2, 3 in the graph whose two arcs connect 1 with 2 and 2 with 3. Therefore the assignment α0 is immediately interpreted as a bicolouring of the vertices of our graph. 3. Given a set of clauses S, a refutation of S is a finite sequence of clauses C1 , C2 , C3 , .

Exercises 39 Exercises 1. 2. 2. Which of the following strings are formulas and which aren’t? (for the latter ones the parsing procedure gets stuck): (((¬XIII ∨ ¬¬XII) ∧ ¬X) ∨ ¬XIII), ¬(¬(¬X ∧ (XIII ∨ X) ∨ XII)), (¬XIII), ((X ∨ X)), (¬XI ∧ XII), ((¬XII ∨ XII)) ∨ XI) ∨ (((¬(¬XIII) ∧ X) ∨ XII)). 3. Prove by induction on the number of connectives: a) in no formula there occurs the sequence of symbols (); b) in each formula the number of open parentheses cannot be smaller than the number of ∨; c) in no formula there occurs the sequence of symbols X¬; d) in each formula there are twice as many parentheses as binary connectives; e) in each formula there are fewer binary connectives than variables; f) if a formula does not have negation symbols and X is its only variable, then it has 4k + 1 symbols, for some k = 0, 1, 2, 3, .

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Logic: a Brief Course by Daniele Mundici


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