If A={1,3,5,7} and B={1,3,7}. Is set B a proper subset of set A? Explain.
Consider the following premises:
1.A-->(B-->A) is a theorem of proportional calculus for all statements A and B.
2.suppose then that the following are the temporary axioms.
a)w
b)y
c)y-->z
Using the logical rules of inference, modus ponens and hypothetical syllogism, show that x-->z is diducible from the given premises.
Let O be the set of odd numbers and O’ = {1, 5, 9, 13, 17, ...} be its subset. Define
the bijections, f and g as:
f : O "\\to" O’, f(d) = 2d - 1, "\\forall" d "\\in" O.
g : "\\Nu" "\\to" O, g(n) = 2n + 1, "\\forall" n "\\isin" "\\Nu" .
Using only the concept of function composition, can there be a bijective map from "\\Nu"
to O’? If so, compute it. If not, explain in details why not.
A Sesotho word cannot begin with of the following letters of alphabet: D, G, V, W,
X, Y and Z.
We define the relation: A Sesotho word x is related to another Sesotho word y if x
begins with the same letter as y.
Determine whether or not this is an equivalence relation.
If it is an equivalence relation then
1. Compute C(sekatana)
2. How many equivalence classes are there in all, and why?
3. What is the partition of the English words under this relation?
If it is NOT an equivalence relation then explain in details why it is not.
Consider the statement form (P↓Q)↓R.
Now, find a restricted statement form logically equivalent to it, in
a) Disjunctive normal form (DNF).
b) Conjunctive normal form (CNF).
Consider the statement form
(P \downarrow Q) \downarrow R
Now, find a restricted statement form logically equivalent to it, in
a) Disjunctive normal form (DNF).
b) Conjunctive normal form (CNF).
Show that ∃xP(x) ∧ ∃xQ(x) and ∃x(P(x) ∧ Q(x)) are not logically equivalent.
Show that ∃x(P(x) ∨ Q(x)) and ∃xP(x) ∨ ∃xQ(x) are logically equivalent.
Determine whether ∀x(P(x) ↔ Q(x)) and ∀x P(x) ↔ ∀xQ(x) are logically equivalent. Justify your answer.
Determine whether ∀x(P(x) → Q(x)) and ∀xP(x) → ∀xQ(x) are logically equivalent. Justify your answer.