Question

What is the correct meaning of the logical expression p→q∨r∧s
?

((p→q)∨r)∧s

p→((q∨r)∧s)

(p→(q∨r))∧s

p→(q∨(r∧s))

Answer #1

Correct answer is **Option (D)
**

**Explanation:**

The operators precedence of is

which means , where "logical and" is greater than "logical or" greater than "implication"

So first we evaluate r and s, then we calculate q or (r and s) and atlast we calculates p implies (q or (r and s))

So the answer is **
**

**Mention in comments if any mistakes are found. Thank
you.**

Prove
a)p→q, r→s⊢p∨r→q∨s
b)(p ∨ (q → p)) ∧ q ⊢ p

p → q, r → s ⊢ p ∨ r → q ∨ s
Solve using natural deduction rules.

Give direct and indirect proofs of:
a. p → (q → r), ¬s ∨ p, q ⇒ s → r.
b. p → q, q → r, ¬(p ∧ r), p ∨ r ⇒ r

Are the statement forms P∨((Q∧R)∨ S) and ¬((¬ P)∧(¬(Q∧ R)∧ (¬
S))) logically equivalent? I found that they were not logically
equivalent but wanted to check. Also, does the negation outside the
parenthesis on the second statement form cancel out with the
negation in front of P and in front of (Q∧ R)∧ (¬ S)) ?

Use a truth table to determine if the following is a
logical equivalence: ( q → ( ¬
q → ( p ∧ r ) ) ) ≡ ( ¬ p ∨ ¬ r )

8. Is the following regular expression property correct?
R* = R*(Ʌ+R)
If it is correct,
prove it. Otherwise, give a counter example to show it is not
correct. (4 points)

8. Is the following regular expression property correct?
R* = R*(Ʌ+R)
If it is correct, prove it. Otherwise, give a counter example to
show it is not correct. (4 points)

1) Show that ¬p → (q → r) and q → (p ∨ r) are logically
equivalent. No truth table and please state what law you're using.
Also, please write neat and clear. Thanks
2) .Show that (p ∨ q) ∧ (¬p ∨ r) → (q ∨ r) is a tautology. No
truth table and please state what law you're using. Also, please
write neat and clear.

Prove or disprove that [(p → q) ∧ (p → r)] and [p→ (q ∧ r)] are
logically equivalent.

Given: (P & ~ R) > (~R & Q), Q> ~P Derive: P >
R. use propositional logic and natural derivation rules.

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