By Eric C.R. Hehner

ISBN-10: 0387941061

ISBN-13: 9780387941066

Realizing programming and programming languages calls for wisdom of the underlying theoretical version. This ebook explores points of programming which are amenable to mathematical facts. the writer describes a programming idea that's a lot less complicated and extra entire than the present theories to this point. within the theoretical version, a specification is simply a boolean expression and refinement is simply a normal implication. the writer develops a pragmatic and vast technique for writing specified standards and designing courses whose executions most likely fulfill the requisites. starting with preparatory fabric in common sense, numbers, units, lists, capabilities and family, the publication advances additional into software idea, the center of the ebook. next chapters could be chosen or passed over in response to path emphasis. The textual content might be priceless to scholars in classes on programming technique or verification on the complicated undergraduate or starting graduate point, in addition to for software program engineers within the box. All technical phrases are defined after which confirmed within the e-book anywhere attainable. No complicated mathematical wisdom or programming language is thought. The e-book comprises a number of routines and worked-out suggestions for particular workouts. Transparency masters and strategies for the remainder workouts can be found from the writer.

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The other three solve the given refinement problem by raising new problems to be solved by further refinement. When these new problems are solved, their solutions will contribute to the solution of the original problem, according to the first of our refinement laws. Refinement by Steps (Stepwise Refinement) (monotonicity, transitivity) If A ⇐ if b then C else D and C ⇐ E and D ⇐ F are theorems, then A ⇐ if b then E else F is a theorem. E is a theorem. If A ⇐ B and B ⇐ C are theorems, then A ⇐ C is a theorem.

Y′ = 2×x Because multiplication has precedence over addition, we must put parentheses around y+1 when we substitute it for x in y′ = 2×x . = y′ = 2×(y+1) (d) x:= 1. x≥1 ⇒ ∃x· y′ = 2×x In x≥1 ⇒ ∃x· y′ = 2×x , the first occurrence of x is nonlocal, and the last occurrence is local. It is the nonlocal x that is being replaced. The local x could have been almost any other name, and probably should have been to avoid any possible confusion. = 1≥1 ⇒ ∃x· y′ = 2×x = even y′ (e) x:= y. x≥1 ⇒ ∃y· y′ = x×y Now we are forced to rename the local y before making the substitution, otherwise we would be placing the nonlocal y in the scope of the local y .

X:= x+y In part (g) we saw that x:= 1. y:= 2 = x′=1 ∧ y′=2 . If we use that, we are then faced with a dependent composition x′=1 ∧ y′=2. x:= x+y for which the Substitution Law does not apply. In a sequence of assignments, it is much better to use the Substitution Law from right to left. = x:= 1. x′ = x+2 ∧ y′=2 = x′=3 ∧ y′=2 (j) x:= 1. if y>x then x:= x+1 else x:= y This part is unremarkable. It just shows that the Substitution Law applies to ifs. = if y>1 then x:= 2 else x:=y (k) x:= 1. x′>x. x′ = x+1 We can use the Substitution Law on the first two pieces of this dependent composition to obtain = x′>1.

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