Download 370/360 Assembler Language Programming by Nancy B. Stern PDF

By Nancy B. Stern

Offers a accomplished business-oriented method of instructing meeting language programming on IBM and IBM-compatible pcs, geared in the direction of freshman and sophomores majoring in information processing or laptop technological know-how. wealthy in pupil aids together with self-evaluation quizzes, bankruptcy goals, workouts, and bankruptcy summaries.

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Extra info for 370/360 Assembler Language Programming

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However, it is not very hard to find a model of the formulas — let the domain be the natural numbers, assign 0 to the constant a and the relation { 1 , 3 , 5 , . e. let p denote the “odd”-relation). g. 1”. Notice that the Herbrand base of a definite program P always is a Herbrand model of the program. To check that this is so, simply take an arbitrary ground instance of any clause A0 ← A1 , . . , Am in P . Clearly, all A0 , . . , Am are in the Herbrand base. Hence the formula is true. However, this model is rather uninteresting — every n-ary predicate of the program is interpreted as the full n-ary relation over the domain of ground terms.

Then in the . k-th iteration the selected equation is of the form Xk = Tk where Ti+1 := f (Ti , Ti ) and T0 := X0 . Hence, |Ti+1 | = 2|Ti | + 1. That is, |Tn | > 2n . This shows the exponential dependency of the unification time on the length of the structures. In this example the growth of the argument lengths is caused by duplication of subterms. As a matter of fact, the same check is repeated many times. Something that could be avoided by sharing various instances of the same structure. In the literature one can find linear algorithms but they are sometimes quite elaborate.

S1 θ = t1 θ, . . , si−1 θ = ti−1 θ, si = ti , si+1 θ = ti+1 θ, . . sn θ = tn θ} (E2 ) where θ := {si /ti }. First assume that σ is a unifier of E1 — that is, sj σ = tj σ for every 1 ≤ j ≤ n. In particular, it must hold that si σ = ti σ. Since si is a variable which is not a subterm of ti it must follow that si /ti σ ∈ σ. Moreover, θσ = σ and it therefore follows that σ is a unifier also of E2 . Next, assume that σ is a unifier of E2 . Thus, si /ti σ ∈ σ and θσ = σ which must then be a unifier also of E1 .

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