Mathematical Reasoning
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New answer posted
a month agoContributor-Level 10
A | B | A | A -> B | ||||
T | T | T | T | T | T | T | T |
T | F | T | F | T | T | F | F |
F | T | T | F | F | F | T | F |
F | F | F | F | F | F | T | F |
A
New answer posted
a month agoContributor-Level 10
Negation of (p → q) ∨ (p ∨ q)
~ [ (p → q) ∨ (p ∨ q)]
≡ ~ (p → q) ∧ ~ (p ∨ q)
[∴ ~ (p → q) ≡ p ∧ ~q]
≡ (p ∧ ~q) ∧ (~p ∧ ~q)
= F (contradiction)
New answer posted
a month agoContributor-Level 10
| p | q | p^q | ¬ (p^q) | (¬ (p^q)vq | pvq | p? q | p? (pvq) |
|-|-|-|-|-|-|-|-|
| T | T | T | F | T | T | T | T |
| T | F | F | T | T | T | F | T |
| F | T | F | T | T | T | T | T |
| F | F | F | T | T | F | T | T |
Tautology, Tautology
New answer posted
a month agoContributor-Level 10
Truth table for (p → q) ∧ (q → ~p).
| p | q | p → q | ~p | q → ~p | (p → q) ∧ (q → ~p) |
|-|-|-|-|-|-|
| T | T | T | F | F | F |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | F | T | T | T | T |
The final column is F, T, which is the truth table for ~p.
Therefore, (p → q) ∧ (q → ~p) is equivalent to ~p.
New question posted
a month agoNew question posted
a month agoNew answer posted
a month agoContributor-Level 9
P: n³ - 1 is even, q: n is odd.
The contrapositive of p → q is ~q → ~p.
~q: n is not odd, i.e., n is even.
~p: n³ - 1 is not even, i.e., n³ - 1 is odd.
⇒ "If n is not odd then n³ - 1 is not even"
⇒ "For an integer n, if n is even, then n³ – 1 is odd."
New answer posted
a month agoContributor-Level 10
Negation of x ↔~ y
≡ ~ (x ↔ ~y)
≡ x ↔ ~ (~y)
≡ x ↔ y
≡ (x ∧ y) ∨ (~x ∧ ~y)
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