Verify the correctness of the following program segment to exchange the values of x and y: temp = x y = temp At the beginning of this program segment, x and y have certain values. Thus we may express the actual precondition as x = a and y = b. The desired postcondi- tion is then x = b and y = a. Using the assignment rule, we can work backward from the postcondition to find the earlier assertions (read the following from the bottom to the top). {y = b, x = a} temp = x {y = b, temp = a} {x = b, temp = a} y = temp {x = b, y = a} The first assertion agrees with the precondition; the assignment rule, applied repeatedly, assures us that the program segment is correct.

The two most recent operands are popped in the postfix expression evaluation example whenever an operator is met so that the subexpression may be evaluated. In the subexpression, the first operand popped is regarded as the second operand, and the second operand popped as the first. Give a case study that exemplifies the significance of this component of the solution.

Use DrRacket. Define the procedures below. Write a procedure reverse-digits (n) which returns a number with the digits of n in reverse order. For example, (reverse-digits 397164) → 461793.   Define a predicate same-digits? (a b) which returns #t if the integers a and b contain the same digits. If a digit occurs n times in a, it must also occur n times in b. For example, (same-digits? 133042 420313) → #t, but (same-digits? 133042 42013) → #f.   To solve the problem, one technique would be to add some extra procedures for manipulating digits inside a number. Here are some hints: To find the number of digits in a number, count how many times you need to divide it by 10 until you reach 0. To extract the second digit of 5784: floor( 5784/102 ) mod 10=7. Note that mod in racket is modulo or remainder To assemble the digits 3, 4, and 5 into a number: 3×102+4×101+5×100 = 345 The standard Scheme procedures modulo and floor may prove useful. The solution only needs to consider non-negative…

The following argument is valid.   ((A ∨ D) → W)  (R ↔ (C ∧ T))  (← R → A)  ← (B ∨ C)  ∴ ← (W → C)  Below is a natural deduction proof, but with some of the lines and justifications  missing. Using any of the proof rules (from Ch. 16 & 18 of the text) in the natural  deduction system, fill in the missing lines and justifications to complete the proof  (3 points).  1 ((A ∨ D) → W) Pr.  2 (R ↔ (C ∧ T)) Pr.  3 (- R → A) Pr.  4 -(B ∨ C) Pr. /∴← (W → C)  5 (- B ∧ - C) _________  6 (W → C) _________  7 _________ ∧E, 5  8 - W MT, 6, 7  9 - (A ∨ D) _________  10 _________ DM, 9  11 - A ∧E, 10  12 _________ MT, 4  13 R _________  14 (C ∧ T) _________  15 C ∧E, 14  16 _________ -E, 7, 15  17 - (W → C) _________