A++ solution
123mathtutorUse De Morgan's laws to find the negation of the following statement.
Kwame will take a job in industry or go to graduate school.
Kwame will not take a job in industry and will not go to graduate school. | |||
Kwame will take a job in industry and will not go to graduate school. | |||
Kwame will not take a job in industry and will go to graduate school. | |||
Kwame will take a job in industry and will go to graduate school. | |||
Kwame will not take a job in industry or will not go to graduate school. |
Complete the truth table for each of these compound propositions.
a) p→(¬q∧r)
p | q | r | ¬q | ¬q∧r | p→(¬q∧r) |
T | T | T | |||
T | T | F | |||
T | F | T |
| ||
T | F | F | |||
F | T | T | |||
F | T | F | |||
F | F | T | |||
F | F | F |
b) ¬p→(q→r)
p | q | r | ¬p | q→r | ¬p→(q→r) |
T | T | T | |||
T | T | F | |||
T | F | T | |||
T | F | F | |||
F | T | T | |||
F | T | F | |||
F | F | T | |||
F | F | F |
c) (p→q)∧(¬p→r)
p | q | r | ¬p | p→q | ¬p→r | (p→q)∧(¬p→r) |
T | T | T | ||||
T | T | F | ||||
T | F | T | ||||
T | F | F | ||||
F | T | T |
| |||
F | T | F | ||||
F | F | T | ||||
F | F | F |
d) (p↔q)∧(¬q↔r)
p | q | r | ¬q | p↔q | ¬q↔r | (p↔q)∧(¬q↔r) |
T | T | T | ||||
T | T | F | ||||
T | F | T | ||||
T | F | F | ||||
F | T | T | ||||
F | T | F | ||||
F | F | T | ||||
F | F | F |
e) (¬p↔¬q)↔(q↔r)
p | q | r | ¬p | ¬q | ¬p↔¬q | q↔r | (¬p↔¬q)↔(q↔r) |
T | T | T | |||||
T | T | F | |||||
T | F | T | |||||
T | F | F | |||||
F | T | T | |||||
F | T | F | |||||
F | F | T | |||||
F | F | F |
Show that each of these conditional statements is a tautology by completing the truth tables.
a) (p∧q)→p
p | q | p∧q | (p∧q)→p |
T | T | ||
T | F | ||
F | T | ||
F | F |
b) q→(p∨q)
p | q | p∨q | q→(p∨q) |
T | T | ||
T | F | ||
F | T | ||
F | F |
c) ¬p→(p→q)
p | q | ¬p | p→q | ¬p→(p→q) |
T | T | |||
T | F | |||
F | T | |||
F | F |
d) (p∧q)→(p→q)
p | q | p∧q | p→q | (p∧q)→(p→q) |
T | T | |||
T | F | |||
F | T | |||
F | F |
e) ¬(p→q)→p
p | q | p→q | ¬(p→q) | ¬(p→q)→p |
T | T | |||
T | F | |||
F | T | |||
F | F |
f) ¬(p→q)→¬q
p | q | p→q | ¬(p→q) | ¬q | ¬(p→q)→¬q |
T | T | ||||
T | F | ||||
F | T | ||||
F | F |
- 10 years ago
Purchase the answer to view it
- a_club_2_-_copy1.pdf