## Exercise 17
Show that ¬(p ↔ q) and p ↔ ¬q are logically equivalent.
A biconditional statement is false if one of its operand is false. Hence p ↔ q is equivalent to ¬p ↔ q and to p ↔ ¬q.
**PERFECT**
## Exercise 29
Show that (p → q) ∧ (q → r) → (p → r) is a tautol-
ogy.
Let's assume thant (p → q) ∧ (q → r) → (p → r) is not a tautology, hence the proposition can be false.
((p → q) ∧ (q → r)) → ((p → r)) to be false (p → q) ∧ (q → r) need to be true AND (p → r) needs to be false.
For (p → r) to be false we need r to be false and p to be true. We can modify the first part of the proposition as follow:(T → q) ∧ (q → F).
For (T → q) ∧ (q → F) to be true we need both (T → q) AND (q → F) to be true.
For (T → q) to be true we need q to be true and for (q → F) to be true we need q to be false.
q can't be true and false at the same time. This is a contradiction therefore (p → q) ∧ (q → r) → (p → r) cannot be false and is a tautology.
**PERFECT**
## Exercise 31
Show that (p → q) → r and p → (q → r) are not logically equivalent.
To find that the proposition is not logically equivalent we need to find value for p, q and r such as the truth values of (p → q) → r and p → (q → r) differ.
If p is false, q is false and r is false, hence (F → F) → F is equivalent to T → F and is false. However F → (F → F) is equivalent to F → T and is true.
Therefore (p → q) → r and p → (q → r) are not logically equivalent.
**PERFECT**
## Exercise 35
Find the dual of each of these compound propositions.
a) p ∧ ¬q ∧ ¬r
b) (p ∧ q ∧ r) ∨ s
c) (p ∨ F) ∧ (q ∨ T)
a) p ∨ ¬q ∨ ¬r
b) (p ∨ q ∨ r) ∧ s
c) (p ∧ T) ∨ (q ∧ F)
**PERFECT**
## Exercise 15
Determine whether (¬q ∧ (p → q)) → ¬p is a tautology.
(¬q ∧ (p → q)) needs to be true and ¬p needs to be false since a conditional statement is false if and only if it has the form T → F.
For (¬q ∧ (p → q)) to be true, p → q needs to be true as well as ¬q.
For (p → q) to be true, p can be false but in this case ¬p would be true which makes (¬q ∧ (p → q)) false.
The only other solution for p → q to be true is that p and q are true, but in that case (¬q ∧ (p → q)) would be again false which doesn't satisfy (¬q ∧ (p → q)) → ¬p is of form T -> F.
Therefore we can conclude that (¬q ∧ (p → q)) → ¬p is a tautology.
**PERFECT**
## Exercise 19
Show that¬p ↔ q and p ↔ ¬q are logically equivalent.
For a biconditional statement to be true both p and q needs to have the same truth value. Since p is negated in the first statement and q in the second, p and q will never have the same truth value. When p is true and q is false both statements will be true, when p and q are true both statements will be false. Therefore both statements are logically equivalent.
** PERFECT **
## Exercise 23
Show that (p → r) ∧ (q → r) and (p ∨ q) → r are logically equivalent.
For a conditional statement to be false it has to be of the form T → F.
For (p → r) ∧ (q → r) to be false both (q → r) and (p → r) needs to be false, that is r needs to be false.
Since (p ∨ q) → r is a conditional statement as well, (p ∨ q) → r will be false since r is false.
For (p → r) ∧ (q → r) to be true there are two solutions:
1. p and q are false whatever the truth value of r
2. p and q are true and r is true as well
For the first case it means that (p ∨ q) is false which makes (p ∨ q) → r true whatever the truth value of r.
For the second case (p ∨ q) → r will be true as well.
We can therefore conclude that (p → r) ∧ (q → r) and (p ∨ q) → r are logically equivalent.
## Exercise 21
Show that ¬(p ↔ q) and ¬p ↔ q are logically equivalent.
We know that p and q needs to be both true or both false for a biconditional statement to be true.
If p and q have the same truth value, ¬(p ↔ q) will be false as ¬p ↔ q (p in that case has another truth value than q)
If p and q don't have the same truth value, in that case ¬(p ↔ q) will be true and negating p but not q will bring p as the same truth value as q. Therefore ¬p ↔ q will be true.
We can conclude that ¬(p ↔ q) and ¬p ↔ q are logically equivalent.
**PERFECT**