Which condition is equivalent to !(a 0 && b 0)?

Answer: a <= 0 || b <= 0 By De Morgan's law, !(P && Q) becomes !P || !Q, and !(a 0) is a <= 0.

Answer: true By De Morgan's law, !(A && B) is equivalent to !A || !B. So p and q are logically equivalent and both evaluate to false here, making p == q true.

"If a then b" can be written as

Answer: !a || b <codea → b</code is logically equivalent to <code(NOT a) OR b</code.

AP Computer Science A · Topic 3.6

Equivalent Boolean Expressions (De Morgan's Laws) Practice

Q: Which condition is equivalent to !(a 0 && b 0)?

Answer: a <= 0 || b <= 0 By De Morgan's law, !(P && Q) becomes !P || !Q, and !(a 0) is a <= 0.

Q: What is printed?

Answer: true By De Morgan's law, !(A && B) is equivalent to !A || !B. So p and q are logically equivalent and both evaluate to false here, making p == q true.

Q: "If a then b" can be written as

Answer: !a || b <codea → b</code is logically equivalent to <code(NOT a) OR b</code.

Part of Boolean Expressions and if Statements.

Practice questions

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Sample questions

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Sample 1difficulty 3/5
```
if (!(a > 0 && b > 0)) {
    System.out.println("no");
}
```
$Which condition is equivalent to !(a > 0 && b > 0)?$
- A
  $!(a > 0) && !(b > 0)$
- B
  $a < 0 && b < 0$
- C
  $a <= 0 || b <= 0$
  check_circle
- D
  $a <= 0 && b <= 0$
Why
$By De Morgan's law, !(P && Q) becomes !P || !Q, and !(a > 0) is a <= 0.$
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Sample 2difficulty 3/5
```
int a = 5;
int b = 10;
boolean p = !(a > 3 && b < 20);
boolean q = (a <= 3 || b >= 20);
System.out.println(p == q);
```
$What is printed?$
- A
  $true$
  check_circle
- B
  $false$
- C
  $p$
- D
  $q$
Why
$By De Morgan's law, !(A && B) is equivalent to !A || !B. So p and q are logically equivalent and both evaluate to false here, making p == q true.$
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Sample 3difficulty 4/5
$"If a then b" can be written as$
- A
  $a || b$
- B
  $a && b$
- C
  $!(a || b)$
- D
  $!a || b$
  check_circle
Why
$<code> a \to b </code> is logically equivalent to <code> (NOT a) OR b </code> .$
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Sample 4difficulty 4/5
$<code> (a || !a) </code> evaluates to$
- A
  $Compile error$
- B
  $false (always)$
- C
  $depends on a$
- D
  $true (always; tautology)$
  check_circle
Why
$Either a or its negation is true; always true.$
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Sample 5difficulty 4/5
```
int a = 5;
int b = 3;
boolean c1 = !(a > 0 || b > 0);
boolean c2 = (a <= 0 && b <= 0);
System.out.println(c1 == c2);
```
$What is printed?$
- A
  $false$
- B
  $true$
  check_circle
- C
  $c1$
- D
  $c2$
Why
$By De Morgan's: !(A || B) is equivalent to !A && !B. So c1 and c2 are logically equivalent (both false here), making c1 == c2 true.$
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