Logic and Computation: Interactive Proof with Cambridge LCF by Lawrence C. Paulson PDF

By Lawrence C. Paulson

ISBN-10: 0521395607

ISBN-13: 9780521395601

Common sense and Computation is anxious with strategies for formal theorem-proving, with specific connection with Cambridge LCF (Logic for Computable Functions). Cambridge LCF is a working laptop or computer application for reasoning approximately computation. It combines tools of mathematical good judgment with area concept, the foundation of the denotational method of specifying the that means of statements in a programming language. This e-book includes components. half I outlines the mathematical preliminaries: uncomplicated good judgment and area thought. they're defined at an intuitive point, giving references to extra complex analyzing. half II offers sufficient element to function a reference guide for Cambridge LCF. it's going to even be an invaluable advisor for implementors of alternative courses according to the LCF procedure.

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2. 2. The functions y = 2x+4 and y = arctan(x) are one-to-one, while y = x2 and y = sin(x) are not. For example: 42 = (−4)2 = 16. 3. The functions y = 2x + 4 and y = x2 are onto, while y = sin(x) and y = arctan(x)are not. For example: the function y = sin(x) does not attain values that are greater than 1. 4. Prove the following lemma. 11 Let f ∶ A → B denote a function. Let A′ ⊆ A and B ′ = B ∖ {f (a′ ) ∈ B ∣ a′ ∈ A∖A′ }. Let g ⊆ A′ ×B ′ denote the relation {(x, f (x)) ∣ x ∈ A′ }. The relation g is a function.

A disk may not be added to a stack if it is larger than the disk currently at the top of the stack. Solving this puzzle with one or two disks is trivial. With three disks we need seven moves. To promote the puzzle, a legend was invented about priests solving a puzzle with k = 64 disks, with the danger that once solved, the world will end. Should we really worry about the truthfulness of the legend? Let fk denote the minimum number of moves required to solve the puzzle with k disks. Formulate a closed formula for fk and prove it by induction.

In Sec. 1, we considered the series ∑ni=1 i. We also proved a formula for this sum. We now consider general arithmetic sequences. Note that the following theorem indeed generalizes Eq. 1 since a0 = 0 and d = 1 in the sequence an = n. 1 Let △ an = a0 + n ⋅ d, n Sn = ∑ ai . △ i=0 Then, Sn = a0 ⋅ (n + 1) + d ⋅ n ⋅ (n + 1) . 1) 36 CHAPTER 3. SEQUENCES AND SERIES Proof: The proof is by induction on n. The induction hypothesis, for n = 0, is easy since S0 = a0 . 1 holds for n. We now prove the induction step.

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Logic and Computation: Interactive Proof with Cambridge LCF by Lawrence C. Paulson

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