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1 Discrete Structures Lecture 25 Quantification IV

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2 Recall: (8.14) Axiom, One Point Rule ( x | x=E : P) = P[x:=E] Provided ¬occurs('x','E') (8.16) Axiom, Range Split (no overlap) ( x | R V S:P) = ( x | R:P) ( x | S:P) provided R S false (8.17) Axiom, Range Split (with overlap) ( x | R V S:P) ( x | R S:P) =( x | R:P) ( x | S:P)

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3 Axioms re: dummies (8.19) Axiom, Interchange of dummies ( x | R:( y | Q:P)) = ( y | Q:( x | R:P) --provided ¬occurs('y','R') and ¬occurs('x','Q') Nested quantifications with the same operator can be interchanged.

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4 Axioms re: dummies (8.20) Axiom, Nesting ( x,y | R Q:P)) = ( x | R:( y | Q:P) --provided ¬occurs('y','R') Single quantification over a list of dummies can be viewed as a nested quantification.

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5 Axioms re: dummies (8.21) Axiom, Dummy Renaming ( x | R:P) = ( y | R[x:=y]:P[x:=y]) --provided ¬occurs('y','R,P') Can replaced dummies with a "fresh" variable as long as it's done consistently.

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6 Splitting Ranges (8.23) Theorem Split off Term ( i | 0 ≤ i < n+1:P) = ( i | 0 ≤ i < n:P) P[i:=n] ( i | 0 ≤ i < n+1:P) = P[i:=0] ( i | 0 < i < n+1:P) NN For n: N and dummies i: N

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7 (8.5) Prove the following theorems. Provided 0 ≤ n (b) ( i | 0 ≤ i ≤ n: b[i]) = ( i | 0 ≤ i < n: b[i]) + b[n] ( i | 0 ≤ i ≤ n: b[i]) = ( i | 0 ≤ i < n: b[i]) + b[n]

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8 Prove 8.5(b) without (8.23), part one (b) ( i | 0 ≤ i ≤ n: b[i]) = ( i | 0 ≤ i < n: b[i]) + b[n] 0 ≤ i ≤ n = 0 ≤ i i ≤ n = 0 ≤ i (i < n V i=n) = (0 ≤ i i<n) V (0 ≤ i i=n) = (0 ≤ i < n) V (0 ≤ i i=n) = (0 ≤ i<n) V (0 ≤ n i=n) = (0 ≤ i < n) V i=n

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9 Now we can prove the theorem (b) ( i | 0 ≤ i ≤ n: b[i]) = ( i | 0 ≤ i < n: b[i]) + b[n] ( i | 0 ≤ i ≤ n: b[i]) = ( i | 0 ≤ i < n V i=n: b[i]) = ( i | 0 ≤ i < n: b[i]) + ( i | i=n: b[i]) = ( i | 0 ≤ i < n: b[i]) + b[n]

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10 Splitting ranges (8.24) (8.24): b ≤ c ≤ d (b ≤ i < d b ≤ i < c V c ≤ i < d) Example: Suppose 0 ≤ n ( V i | 0 ≤ i < n+1: b[i]=0) = <(8.24) b,c,d:=0,n,n+1 0 ≤ n ≤ n+1 (0 ≤ i ( V i | 0 ≤ i < n V n ≤ i < n+1: b[i]=0) Now we can split the range.

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11 8.6(a) without using 8.23 ( V i | 0 ≤ i < n+1: b[i]=0) ( V i | 0 ≤ i < n: b[i]=0) V b[n]=0 ( V i | 0 ≤ i < n+1: b[i]=0) = ( V i | 0 ≤ i < n V n ≤ i < n+1: b[i]=0) = ( V i | 0 ≤ i < n: b[i]=0) V ( V i | n ≤ i < n+1: b[i]=0) = ( V i | 0 ≤ i < n: b[i]=0) V b[n]=0

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12 8.6(c) without using 8.23, part one ( V i | 0 ≤ i < n+1: b[i]=0) b[0]=0 V ( V i | 0 < i < n+1: b[i]=0) 0 ≤ i < n+1 = 0 ≤ i i < n+1 = (0=i V 0<i) i < n+1 = (0 = i i<n+1) V (0 < i i<n+1) = (0 = i i<n+1) V (0 < i<n+1) = (0 = i 0<n+1) V (0 < i<n+1) 0 = i V (0 < i<n+1)

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13 8.6(c) without using 8.23 ( V i | 0 ≤ i < n+1: b[i]=0) b[0]=0 V ( V i | 0 < i < n+1: b[i]=0) ( V i | 0 ≤ i < n+1: b[i]=0) = ( V i | 0 = i V 0 < i < n+1: b[i]=0) = ( V i |0 = i: b[i]=0) V ( V i | 0 < i < n+1: b[i]=0) = b[0]=0 V ( V i |0 < i < n+1: b[i]=0)

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14 A Note about ranges For example: 2, …, 15 (a) 2 ≤ i ≤ 15 (b) 2 ≤ i < 16 (c) 1 < i ≤ 15 (d) 1 < i < 16 Number of elements in range is equal to upperBound - lowerBound in (b) & (c). Xerox Parc study, said fewest errors with (b).

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