MacsNDJ

1. Kumera Nemomsa
Part I
a).Let's evaluate the statement for each possible combination:
For x = 1:
For y = 1: 1^2 + 2(1) = 3 < 10 (True)
For y = 2: 1^2 + 2(2) = 5 < 10 (True)
For y = 3: 1^2 + 2(3) = 7 < 10 (True)
For x = 2:
For y = 1: 2^2 + 2(1) = 6 < 10 (True)
For y = 2: 2^2 + 2(2) = 10 < 10 (False)
For y = 3: 2^2 + 2(3) = 14 < 10 (False)
Since we found some combinations of x and y where the statement is false, the overall truth
value of (∀x)(∀y)(x^2+2y<10) is false.
In other words, there exist values of x and y within the given set u = {1, 2, 3} for which the
expression x^2 + 2y is greater than or equal to 10.
b).Let's evaluate the statement for each value of x:
For x = 1:
For y = 1: 1^2 + 2(1) = 3 < 10 (True)
For y = 2: 1^2 + 2(2) = 5 < 10 (True)
For y = 3: 1^2 + 2(3) = 7 < 10 (True)

2. Since the inequality holds true for all values of y when x = 1, we can conclude that (∃x)(∀y)(x^2
+ 2y < 10) is true for x = 1.
Therefore, the statement (∃x)(∀y)(x^2 + 2y < 10) is true for the given set u = {1, 2, 3} because
there exists at least one value of x (in this case, x = 1) for which the inequality holds true for all
values of y within the set.
c).
Let's evaluate the statement for each value of x:
For x = 1:
For y = 1: 1^2 + 2(1) = 3 < 10 (True)
For y = 2: 1^2 + 2(2) = 5 < 10 (True)
For y = 3: 1^2 + 2(3) = 7 < 10 (True)
Since the inequality holds true for all values of y when x = 1, we can conclude that (∃x)(∀y)(x^2
+ 2y < 10) is true for x = 1.
Therefore, the statement (∃x)(∀y)(x^2 + 2y < 10) is true for the given set u = {1, 2, 3} because
there exists at least one value of x (in this case, x = 1) for which the inequality holds true for all
values of y within the set.
d).
Let's evaluate the statement for each combination of x and y:
For x = 1:
For y = 0: 1^2 + 2(0) = 1 < 10 (True)
For y = 1: 1^2 + 2(1) = 3 < 10 (True)
For y = 2: 1^2 + 2(2) = 5 < 10 (True)
For y = 3: 1^2 + 2(3) = 7 < 10 (True)
For x = 2:
For y = 0: 2^2 + 2(0) = 4 < 10 (True)
For y = 1: 2^2 + 2(1) = 6 < 10 (True)
For y = 2: 2^2 + 2(2) = 8 < 10 (True)
For y = 3: 2^2 + 2(3) = 10 < 10 (False)

3. Since there exist values of x and y within the set u = {1, 2, 3} for which the inequality x^2 + 2y <
10 holds true, the statement (∃x)(∃y)(x^2 + 2y < 10) is true.
In other words, there exists at least one pair of values (x, y) within the set u such that the
expression x^2 + 2y is less than 10, which means the statement (∃x)(∃y)(x^2 + 2y < 10) is true.
Part II
Prove the Validity of the argument
a).
To prove the validity of the argument:
Premise 1: (p → q) ∧ (r → s)
Premise 2: p ∨ r
Conclusion: ¬q → s
We can use deductive reasoning and logical rules to demonstrate the validity of the argument.
Here's a proof:
Assume ¬q (Assumption for Conditional Proof)
Assume ¬s (Assumption for Conditional Proof)
From premise 1, (p → q) ∧ (r → s), we have p → q (Simplification)
From premise 2, p ∨ r, we have two cases to consider:
4.1. Case 1: Assume p
4.1.1. From ¬q (assumption), we derive a contradiction: ¬q and q (Contradiction Introduction)
4.1.2. As a contradiction has been reached, we can conclude that the assumption p is false
(Proof by contradiction)
4.2. Case 2: Assume r
4.2.1. From r → s (premise 1) and r (assumption), we can conclude s (Modus Ponens)

4. Since in both cases, either p or r leads to a contradiction, we can conclude that both p and r are
false (Disjunctive Syllogism)
Since p is false, we can conclude ¬p (Negation Introduction)
Since ¬p and ¬q, we have ¬p ∧ ¬q (Conjunction Introduction)
Using De Morgan's law, we can rewrite the conjunction as ¬(p ∨ q) (De Morgan's Law)
From ¬(p ∨ q), we can derive p ∨ q (Double Negation)
From p ∨ q, we have q (Disjunctive Syllogism)
From q and ¬q, we have a contradiction: q and ¬q (Contradiction Introduction)
As a contradiction has been reached, our initial assumption ¬q must be false (Proof by
contradiction)
Therefore, we can conclude ¬q → s (Conditional Proof: ¬q → s)
Therefore, the argument is valid as we have shown that the conclusion ¬q → s logically follows
from the premises.
By going through this proof, we have demonstrated that if the premises (p → q) ∧ (r → s) and p
∨ r hold, then the conclusion ¬q → s is necessarily true. Hence, the argument is valid.
b).
To prove the validity of the argument:
Premise 1: p → q
Premise 2: ¬p → r
Premise 3: r → s
Conclusion: ¬q → s
We can use deductive reasoning and logical rules to demonstrate the validity of the argument.
Here's a proof:
Assume ¬q (Assumption for Conditional Proof)
From premise 1, p → q, and the assumption ¬q, we can conclude ¬p (Modus Tollens)
From premise 2, ¬p → r, and ¬p (derived from step 2), we can conclude r (Modus Ponens)
From premise 3, r → s, and r (derived from step 3), we can conclude s (Modus Ponens)
Therefore, from the assumption ¬q, we have derived s

5. Thus, ¬q → s (Conditional Proof)
Therefore, the argument is valid as we have shown that the conclusion ¬q → s logically follows
from the premises.
By going through this proof, we have demonstrated that if the premises p → q, ¬p → r, and r →
s hold, then the conclusion ¬q → s is necessarily true. Hence, the argument is valid.
c).
To prove the validity of the argument:
Premise 1: (¬p ∧ ¬q)
Premise 2: (q ∨ r) → p
Conclusion: ¬r
We can use deductive reasoning and logical rules to demonstrate the validity of the argument.
Here's a proof:
Assume r (Assumption for Conditional Proof)
From Premise 2, we have (q ∨ r) → p
Since we assumed r, we can conclude (q ∨ r) (Addition)
From (q ∨ r) → p and (q ∨ r) (derived from steps 2 and 3), we can conclude p (Modus Ponens)
From Premise 1, we have (¬p ∧ ¬q)
From (¬p ∧ ¬q), we have ¬p (Simplification)
From p and ¬p, we have a contradiction: p and ¬p (Contradiction Introduction)
As a contradiction has been reached, our initial assumption r must be false (Proof by
contradiction)
Therefore, ¬r (Negation Introduction)
Therefore, the argument is valid as we have shown that the conclusion ¬r logically follows from
the premises.
By going through this proof, we have demonstrated that if the premises (¬p ∧ ¬q) and (q ∨ r) →
p hold, then the conclusion ¬r is necessarily true. Hence, the argument is valid.
d).

6. To prove the validity of the argument:
Premise 1: ¬r ∧ ¬s
Premise 2: (¬s → p) → r
Conclusion: ¬p
We can use deductive reasoning and logical rules to demonstrate the validity of the argument.
Here's a proof:
Assume p (Assumption for Conditional Proof)
From Premise 2, (¬s → p) → r, and the assumption p, we can conclude r (Modus Ponens)
From Premise 1, ¬r ∧ ¬s, we have ¬r (Simplification)
From ¬r and r, we have a contradiction: ¬r and r (Contradiction Introduction)
As a contradiction has been reached, our initial assumption p must be false (Proof by
contradiction)
Therefore, ¬p (Negation Introduction)
Therefore, the argument is valid as we have shown that the conclusion ¬p logically follows from
the premises.By going through this proof, we have demonstrated that if the premises ¬r ∧ ¬s
and (¬s → p) → r hold, then the conclusion ¬p is necessarily true. Hence, the argument is valid.