1 and 2 I mean, what about 3 4 5

Mathematics

ScanSolve AI Answer

Step-by-step solution

9 steps

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 \lor Q) \Rightarrow (P \land Q)
PPQQP∨QP \lor QP∧QP \land Q(P∨Q)⇒(P∧Q)(P \lor Q) \Rightarrow (P \land Q)
TTTTT
TFTFF
FTTFF
FFFFT
  • b) (P⇒Q)⇔(∼Q⇒∼P)(P \Rightarrow Q) \Leftrightarrow (\sim Q \Rightarrow \sim P)
PPQQ∼P\sim P∼Q\sim QP⇒QP \Rightarrow Q∼Q⇒∼P\sim Q \Rightarrow \sim P(P⇒Q)⇔(∼Q⇒∼P)(P \Rightarrow Q) \Leftrightarrow (\sim Q \Rightarrow \sim P)
TTFFTTT
TFFTFFT
FTTFTTT
FFTTTTT

4) Using truth tables, verify whether the following arguments are valid or invalid:

  • a) P⇒QP \Rightarrow Q ∼P\sim P ∴∼Q\therefore \sim Q

This argument is invalid.

PPQQP⇒QP \Rightarrow Q∼P\sim P∼Q\sim Q
TTTFF
TFFFT
FTTTF
FFTTT
  • Explanation: An argument is valid if, whenever all premises are true, the conclusion is also true. In the third row, both premises (P⇒QP \Rightarrow Q and ∼P\sim P) are true, but the conclusion (∼Q\sim 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∨QP \lor Q ∼P\sim P ∴Q\therefore Q

This argument is valid.

PPQQP∨QP \lor Q∼P\sim PQQ
TTTFT
TFTFF
FTTTT
FFFTF
  • Explanation: We look for rows where all premises (P∨QP \lor Q and ∼P\sim P) are true. This occurs only in the third row. In this row, the conclusion (QQ) 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)\sim (P \lor \sim Q) \lor (\sim P \land \sim Q)
  1. Apply De Morgan's Law to the first term.

    ∼(P∨∼Q)≡(∼P∧∼(∼Q))≡(∼P∧Q)\sim (P \lor \sim Q) \equiv (\sim P \land \sim (\sim Q)) \equiv (\sim P \land Q)
  2. Substitute back into the original expression.

    (∼P∧Q)∨(∼P∧∼Q)(\sim P \land Q) \lor (\sim P \land \sim Q)
  3. Apply the Distributive Law.

    ∼P∧(Q∨∼Q)\sim P \land (Q \lor \sim Q)
  4. Apply the Complement Law (Q∨∼Q≡TQ \lor \sim Q \equiv T).

    ∼P∧T\sim P \land T
  5. Apply the Identity Law (∼P∧T≡∼P\sim P \land T \equiv \sim P).

    ∼P\boxed{\sim P}
    • b) (P⇒Q)∧(P⇒∼Q)(P \Rightarrow Q) \land (P \Rightarrow \sim Q)
  6. Replace implications with disjunctions using A⇒B≡∼A∨BA \Rightarrow B \equiv \sim A \lor B.

    (∼P∨Q)∧(∼P∨∼Q)(\sim P \lor Q) \land (\sim P \lor \sim Q)
  7. Apply the Distributive Law.

    ∼P∨(Q∧∼Q)\sim P \lor (Q \land \sim Q)
  8. Apply the Complement Law (Q∧∼Q≡FQ \land \sim Q \equiv F).

    ∼P∨F\sim P \lor F
  9. Apply the Identity Law (∼P∨F≡∼P\sim P \lor F \equiv \sim P).

    ∼P\boxed{\sim P}
Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question