By Manfred Broy, Stefan Jähnichen

This publication constitutes the ultimate file of the paintings conducted within the venture KORSO ("Korrekte Software") funded by means of the German Federal Ministry for learn and expertise. KORSO is an evolutionary, prototype-oriented venture geared toward bettering the theoretical foundations of quality-driven software program engineering and at imposing identified concepts for purposes of useful relevance.
The 21 strictly refereed papers offered are prepared in 5 sections on tools for correctness, languages, improvement structures and logical frameworks, instruments, and case reports. furthermore, the preface and introductory paper provide useful historical past info and a concise state of the art overview.

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Mcreate a Similar reasoning suffices to show that fst is a base map from r0 to r1 and snd is a base map from r0 to r2 , so we can conclude that R and (l , r ) constitute a bisimulation between sp 1 and sp 2 , that is, sp 1 ≡b sp 2 . 10. Given sp 1 :: A then sp 1 ≡s sp 2 . S1 B , sp 2 :: A S2 B , if sp 1 ≡b sp 2 ♦ Proof. For convenience, we again write sp 1 = (l1 , r1 ) and sp 2 = (l2 , r2 ). We are given R and a span sp 0 :: A R B constituting a bisimulation sp 1 ≡b sp 2 . Let sp 0 = (l0 , r0 ). put s2 b) ∈ R 28 F.

Mcreate b We must now show that l0 and r0 are well-behaved (full) lenses, and that the projections fst and snd map sp = (l0 , r0 ) to sp 1 and sp 2 respectively. We first show that l0 is well-behaved; the reasoning for r0 is symmetric. mget (s1 , s2 ))} = [[ Definition ]] 26 F. Abou-Saleh et al. mcreate a; return ((s1 , s2 ), a)} Next, we show that fst is a base map from l0 to l1 and snd is a base map from l0 to l2 . It is easy to show that fst is a base map from l0 to l1 by unfolding definitions and applying of monad laws.

If ml 1 :: [σ1 β]μ and ml 2 :: [σ2 σ2 ]μ . is ml 1 ml 2 :: [σ1 (σ1 σ2 ) β]μ are well-behaved then so ♦ Proof. It suffices to consider the two lenses l1 = MLens fst put L create L and l2 = MLens snd put R create R in isolation. Moreover, the two cases are completely symmetric, so we only show the first. mget s2 )} = [[ (MGetPut) ]] return s The proof for (MPutGet) goes as follows. Note that it holds by construction, without appealing to well-behavedness of ml 1 or ml 2 . mput (s1 , s2 ) a; return ((s1 , s2 ), a)} The proof for (MCreateGet) is similar.

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