By James H. Martin
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The next refinement makes y′=2x in two steps: first y′=2x–1 and then double y . The antecedent x>0 ensures that 2x–1 will be natural. The last two refinements again contain superfluous assignments. Without the theory of programming, we would be very worried that these superfluous assignments might in some way make the result wrong. With the theory, we only need to prove these six refinements, and we are confident that execution will not give us a wrong answer. 4 Program Theory 46 This solution has been constructed to make it difficult to follow the execution.
The limit of the function (limit of the sequence) is expressed as LIM f . 718281828459 . We define the LIM quantifier by the following Limit Axiom: (MAX m· MIN n· f(m+n)) ≤ (LIM f) ≤ (MIN m· MAX n· f(m+n)) 33 3 Function Theory with all domains being nat . This axiom gives a lower bound and an upper bound for LIM f . When those bounds are equal, the Limit Axiom tells us LIM f exactly. For example, LIM n· 1/(n+1) = 0 For some functions, the Limit Axiom tells us a little less. For example, –1 ≤ (LIM n→(–1)n) ≤ 1 In general, (MIN f) ≤ (LIM f) ≤ (MAX f) For monotonic (nondecreasing) f , LIM f = MAX f .
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. x′ = x+1 Now we have to use the axiom for dependent composition to get a further simplification. = ∃x′′, y′′· x′′>1 ∧ x′ = x′′+1 = x′>2 The error we avoided in the first step is to replace x with 1 in the last part of the composition x′ = x+1 . 2 Refinement Two specifications P and Q are equal if and only if each is satisfied whenever the other is.