8: Maximum and Minimum in .NET

Integrating Data Matrix ECC200 in .NET 8: Maximum and Minimum
8: Maximum and Minimum
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Second and third, we use the fact that true is a unit of conjunction and, again, ^ is reflexive by instantiating z to x and z to y we get, respectively,
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x\y ^ x =
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A fourth property can be obtained from the fact that ^ is a total ordering. That is, for all x and y, x^y v y^x . Using this fact, we calculate: x\y = x v x\y = y { = antisymmetry of ^ } (x\y ^ x A x^x\y) v (x\y ^ y A y^xly) { above, substitution of equals for equals } A true) v (x^y A true) { true is the unit of conjunction } v x^y = { true . So maximum is a choice function: x\y chooses between x and y. Here are a couple of very short exercises. Exercise 8.2. Show that x\y ^ x = x\y = x. Exercise 8.3. Simplify the statement x\y ^ x+y. D D ^ is a total ordering }
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8.2 Using Indirect Equality
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To derive additional equalities, we recall the rule of indirect equality: in order to establish the equality x = y , show that, for arbitrary z of the same type as x and
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Here are two examples. In the first example, we show that maximum is associative because conjunction is associative. We have, for all w, definition of max }
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8.2 Using Indirect Equality
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{ { {
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definition of max
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A y^w) A z^w
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A is associative } A (y^w A z^w) definition of max (applied twice) }
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Thus, by indirect equality, (x\y}\z = x\(y\z] . Note how short and straightforward this proof is. In contrast, if maximum is defined in the conventional way by case analysis, it would be necessary to consider six different cases, six being the number of ways to order three values x, y and z. In the second example, we derive a distributivity property of maximum. We have, for all w,
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x + (y\z] ^ w
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{ y\z ^ w-x = = { definition of max } y ^w-x A z ^w-x { shunt 'x +' back in order to be able to apply the definition of maximum } x + y^w A x + z^w { definition of max } (x+y}\(x+z) ^ w . Thus, by indirect equality, x + (y\z] = (x+y)\(x+z] . Try the following for yourself, observing carefully the properties of conjunction that you exploit. Exercise 8.4. Prove the following. (a) x\x = x . (b) x\y = y\x . D = shunt 'x +' out of the way in order to be able to apply the definition of maximum }
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8: Maximum and Minimum Another property we can derive from the definition is by using contraposition: true = = { { { = { definition of maximum } contrapositive } De Morgan } -i(w^v) = v <u } x t y s ^ z = x s=5 z A y s^ z ->(xty ^ z) = - - ( x ^ Z A y ->(xTy ^ z) = ->(x<z) v z<x\y = z<xvz<y . Thus we have derived
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z<x\y = z<xvz<y .
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This too is a distributMty property: the function (z <) distributes over maximum turning it into a disjunction. Now we verify that the function (z ^) also distributes over maximum using the properties of disjunction as opposed to the properties of conjunction. v { { definition } rearrangement of terms (allowed because disjunction is symmetric and associative) }
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(z = x v z = .y) v (z<x v z<y) (z = x v z<x) v (z = y v z<y)
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in the second disjunct: z<x\y = z <x v z <y } See the calculation below. }
z = x v z = y v z<x\y
The last step in the above needs justification. On the face of it, it looks like the step is an easy one. After all, = z = x\y v z<x\y , so that it looks tike we just need to replace z = x v z = y b y z = xty. But it is not the case that z = x v z = y = z = x!yin general. A bit more thought is necessary.