Here are the solutions for questions 3, 4, and 5.
3) Construct a truth table for the following compound propositions:
- a) (P∨Q)⇒(P∧Q)
| P | Q | P∨Q | P∧Q | (P∨Q)⇒(P∧Q) |
| :--: | :--: | :--------: | :--------: | :---------------------------------: |
| T | T | T | T | T |
| T | F | T | F | F |
| F | T | T | F | F |
| F | F | F | F | T |
- b) (P⇒Q)⇔(∼Q⇒∼P)
| P | Q | ∼P | ∼Q | P⇒Q | ∼Q⇒∼P | (P⇒Q)⇔(∼Q⇒∼P) |
| :--: | :--: | :--: | :--: | :---------------: | :-----------------------: | :------------------------------------------------------: |
| T | T | F | F | T | T | T |
| T | F | F | T | F | F | T |
| F | T | T | F | T | T | T |
| F | F | T | T | T | T | T |
4) Using truth tables, verify whether the following arguments are valid or invalid:
- a) P⇒Q
∼P
∴∼Q
This argument is invalid.
| P | Q | P⇒Q | ∼P | ∼Q |
| :--: | :--: | :---------------: | :--: | :--: |
| T | T | T | F | F |
| T | F | F | F | T |
| F | T | T | T | F |
| F | F | T | T | T |
-
Explanation: An argument is valid if, whenever all premises are true, the conclusion is also true. In the third row, both premises (P⇒Q and ∼P) are true, but the conclusion (∼Q) is false. This is a counterexample, showing the argument is invalid. This is an example of the Fallacy of Denying the Antecedent.
-
b) P∨Q
∼P
∴Q
This argument is valid.
| P | Q | P∨Q | ∼P | Q |
| :--: | :--: | :--------: | :--: | :--: |
| T | T | T | F | T |
| T | F | T | F | F |
| F | T | T | T | T |
| F | F | F | T | F |
- Explanation: We look for rows where all premises (P∨Q and ∼P) are true. This occurs only in the third row. In this row, the conclusion (Q) is also true. Therefore, the argument is valid. This is a valid form of argument known as Disjunctive Syllogism.
5) Using the laws of logic, simplify the following compound propositions:
- a) ∼(P∨∼Q)∨(∼P∧∼Q)
Step 1: Apply De Morgan's Law to the first term.
∼(P∨∼Q)≡(∼P∧∼(∼Q))≡(∼P∧Q)
Step 2: Substitute back into the original expression.
(∼P∧Q)∨(∼P∧∼Q)
Step 3: Apply the Distributive Law.
∼P∧(Q∨∼Q)
Step 4: Apply the Complement Law (Q∨∼Q≡T).
∼P∧T
Step 5: Apply the Identity Law (∼P∧T≡∼P).
∼P
- b) (P⇒Q)∧(P⇒∼Q)
Step 1: Replace implications with disjunctions using A⇒B≡∼A∨B.
(∼P∨Q)∧(∼P∨∼Q)
Step 2: Apply the Distributive Law.
∼P∨(Q∧∼Q)
Step 3: Apply the Complement Law (Q∧∼Q≡F).
∼P∨F
Step 4: Apply the Identity Law (∼P∨F≡∼P).
∼P
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