By Sam Lindley, Conor McBride, Phil Trinder, Don Sannella

This quantity is released in Honor of Philip Wadler at the party of his sixtieth birthday, and the gathering of papers shape a Festschrift for him. The contributions are made via the various many that comprehend Phil and feature been prompted by way of him. The study papers integrated the following symbolize a number of the parts within which Phil has been energetic, and the editors thank their colleagues for agreeing to give a contribution to this Festschrift. we try to summarize Phil Wadler's medical achievements. additionally, we describe the non-public type and exuberance that Phil has delivered to the subject.

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Extra resources for A List of Successes That Can Change the World: Essays Dedicated to Philip Wadler on the Occasion of His 60th Birthday

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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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