Relations Given the sets over the domain {a, b, c, d}: A = { (a, a), (b, b), (d, d), (b, a) } B = { (a, c), (c, c), (c, a), (d, a) } C = { (a, c), (a, d), (c, b), (c, c) } For A, B, and C: find the following closures. 1. Reflexive 2. Symmetric 3. Transitive Logic
Relations Given the sets over the domain {a, b, c, d}: A = { (a, a), (b, b), (d, d), (b, a) } B = { (a, c), (c, c), (c, a), (d, a) } C = { (a, c), (a, d), (c, b), (c, c) } For A, B, and C: find the following closures. 1. Reflexive 2. Symmetric 3. Transitive Logic
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Relations
Given the sets over the domain {a, b, c, d}:
A = { (a, a), (b, b), (d, d), (b, a) }
B = { (a, c), (c, c), (c, a), (d, a) }
C = { (a, c), (a, d), (c, b), (c, c) }
For A, B, and C: find the following closures.
1. Reflexive
2. Symmetric
3. Transitive
Logic
1. Simplify the following using a truth table.
¬( (p ∧ q) -> (p ∨ q))
2. Show if the following is equivalent to ¬ p ∧ ¬ q . Use Boolean math. Label each law that you apply.
¬(p ∨ (¬ p ∧ q))
3. Simply the following using Boolean Math. Label each law that you apply.
b ∧ a ∨ e ∧ c ∧ ¬e ∨ ¬a ∧ b
4. Write out and prove (or disprove) the following statement.
“If I’m sleepy, then I’ll drink coffee or tea. I’ll never drink tea. Therefore, if I’m sleepy, then I'll drink coffee"
Convert the English sentence to an argument (short notation). Then, use a truth table to prove/disprove it:
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