Example A Sample Clauses

Example A. If, on December 31, 2022, at least 3,030 rooms have been rebranded, there is no change to the Diamond Converted Brand Royalty on January 1, 2023 and it will remain at 2.00% until September 1, 2023, when it changes to 3% in accordance with the Baseline Fee Schedule. If, on December 31, 2022, less than 3,030 rooms but at least 2,652 rooms are rebranded, then starting on January 1, 2023, the Adjusted Diamond Converted Brand Royalty shall increase to 2.50% (instead of 2.00%) in accordance with Threshold 1 Schedule. Accordingly, in such case, on September 1, 2023, the Adjusted Diamond Converted Brand Royalty shall increase to 3.50% (instead of the normal increase to 3.00%) in accordance with Threshold 1 Schedule. If on December 31, 2022, less than 2,652 rooms have been rebranded, then starting on January 1, 2023, the Adjusted Diamond Converted Brand Royalty shall increase to 3.00% (instead of 2.0%) in accordance with Threshold 2 Schedule. Accordingly, in such case, on September 1, 2023, the Adjusted Diamond Converted Brand Royalty fee shall increase to 4.00% (instead of the normal increase to 3.00%) in accordance with Threshold 2 Schedule.

Example A. This is the first of three worked examples. Consider a transformation between two classes A and B, in which each object x of A is transformed into an object y of B, subject to three conditions (see Figure 4.2). • The value of attribute B1 of y is the same as the value of attribute A1 of x. • The value of attribute B2 of y is the same as the value of attribute A3 of x, if the value of attribute A2 of x is True. • The value of attribute B2 of y is the same as the value of attribute A4 of x, if the value of attribute A2 of x is False. A1 : Nat f A2 : Bool A3 : Nat A4 : Nat B B1 : Nat B2 : Nat The logical interpretation of the transformation is given by ∀x : A. Iy : B . P (x, y) , where P (x, y) is the conjunction of three propositions, i.e. , , z4.1 r (A1 x = B1 y) ∧ (A2 x = True → A3 x = B2 y) ∧ (A2 x = False → A4 x = B2 y) . The introduction rules for types A and B are as follows. n1 : Nat b : Bool n2 : Nat n3 : Nat (AI) . n1 : Nat n2 : Nat (B I) . @A n1 b n2 n3 : A @B n1 n2 : B The elimination and computation rules are intentionally undefined. Further, the attribute functions A1 through A4, and B1 and B2, are as usual defined by the destructors @—1 and , , A @—1 of A and B respectively. Four steps are required to prove that z4.1 ris inhabited and that the transformation is certifiable as follows. 1. Prove [x : A] ▶ (A1 x = B1 y) [(f x)/y] , for a suitable function f of type A → B. Now, in virtue of its type, f must be a function of the form λx. @B n1 n2 for some n1 and n2, where n1 and n2 determine the values assigned to attributes B1 and B2 respectively. Note that n1 and n2 both depend on x. Clearly, n1 must be (A1 x), and n2 must be an “if-then-else” term, which is conditional on the value of attribute A2, i.e. f =df λx. @B (A1 x) (if (A2 x) then (A3 x) else (A4 x)) . The proof follows by the (=exp) rule, since (A1 x = B1 y) [(f x)/y] → A1 x = B1 (f x) → A1 x = A1 x ,

Example A. ▇▇. ▇▇▇▇ ▇▇▇ is an 8 - hour employee and has chargeable absences totaling 21.8 hours.

Example A. A teacher employed on a 20% contract who has earned 3 SLDs takes 3 days at 20%, which equals 0.6 days from the bank of 3. 2.4 days remain in the bank to be carried forward or paid out.