###### UGC NET CS 2014 June-paper-3

October 4, 2023###### Programming

October 4, 2023# Engineering-Mathematics

Question 5 |

Let p, q and s be four primitive statements. Consider the following arguments:

P: [(¬p ∨ q) ∧ (r → s) ∧ (p ∨ r)] → (¬s → q)

Q: [(¬p ∧ q) ∧ [q → (p → r)] → ¬r

R: [[(q ∧ r) → p] ∧ (¬q ∨ p)] → r

S: [p ∧ (p → r) ∧ (q ∨ ¬r)] → q

Which of the above arguments are valid?

P and Q only | |

P and R only | |

P and S only | |

P, Q, R and S |

Question 5 Explanation:

Let p→q is conditional proposition here. p and q are compound propositions itself. Arguments to be valid if all combinations have to be tautology (like T→T, F→T, F→F) and its invalid if it have fallacy (T→F).

If we somehow get this fallacy (T→F) then an argument is invalid.

For options P and S you don’t get any such combinations for T→F, so P and S are valid.

For option Q: If we put p=F, q=T, r=T then we get T→F. So its INVALID.

For option R: If we put p=F, q=F, r=F then we get T→F. So it is INVALID.

So, answer is (C).

If we somehow get this fallacy (T→F) then an argument is invalid.

For options P and S you don’t get any such combinations for T→F, so P and S are valid.

For option Q: If we put p=F, q=T, r=T then we get T→F. So its INVALID.

For option R: If we put p=F, q=F, r=F then we get T→F. So it is INVALID.

So, answer is (C).

Correct Answer: C

Question 5 Explanation:

Let p→q is conditional proposition here. p and q are compound propositions itself. Arguments to be valid if all combinations have to be tautology (like T→T, F→T, F→F) and its invalid if it have fallacy (T→F).

If we somehow get this fallacy (T→F) then an argument is invalid.

For options P and S you don’t get any such combinations for T→F, so P and S are valid.

For option Q: If we put p=F, q=T, r=T then we get T→F. So its INVALID.

For option R: If we put p=F, q=F, r=F then we get T→F. So it is INVALID.

So, answer is (C).

If we somehow get this fallacy (T→F) then an argument is invalid.

For options P and S you don’t get any such combinations for T→F, so P and S are valid.

For option Q: If we put p=F, q=T, r=T then we get T→F. So its INVALID.

For option R: If we put p=F, q=F, r=F then we get T→F. So it is INVALID.

So, answer is (C).

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