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Chapter 2: Problem 19

For any statements \(p, q\), prove that a) \(\neg(p \downarrow q) \Leftrightarrow(\neg p \uparrow \neg q)\) b) \(\neg(p \uparrow q) \Leftrightarrow(\neg p \downarrow \neg q)\)

Short Answer

Expert verified
By applying De Morgan's laws, the equivalences \(\neg(p \downarrow q) \Leftrightarrow (\neg p \uparrow \neg q)\) and \(\neg(p \uparrow q) \Leftrightarrow (\neg p \downarrow \neg q)\) are proven to be correct.

Step by step solution

01

Understanding the Problem: Logical Operators

Firstly, it's essential to understand the symbols being used. Here, the down-arrow ( \(\downarrow\) ) represents NOR (a negated OR operation), meanwhile the up-arrow ( \(\uparrow\) ) represents NAND (a negated AND operation). The tilde ( \(\neg\) symbol represents negation.
02

Part a: Proof of \(\neg(p \downarrow q) \Leftrightarrow (\neg p \uparrow \neg q)\)

To prove this, let's start from the left-hand side. \(\neg(p \downarrow q)\) can be re-written as \(\neg(\neg (p \vee q))\), since \((p \downarrow q)\) is equivalent to \( \neg (p \vee q)\). According to De Morgan's laws, the above expression can be rewritten as \((\neg p \wedge \neg q)\). Finally, \((\neg p \wedge \neg q)\) equals to \((\neg p \uparrow \neg q)\), which is the right-hand side. Therefore, we proved that \(\neg(p \downarrow q) \Leftrightarrow (\neg p \uparrow \neg q)\).
03

Part b: Proof of \(\neg(p \uparrow q) \Leftrightarrow (\neg p \downarrow \neg q)\)

Similarly as in part a, start from the left-hand side. \(\neg(p \uparrow q)\) is interpreted as \(\neg(\neg (p \wedge q))\), due to the definition of NAND operation - the up-arrow signifies \((p \uparrow q)\). According to De Morgan's laws, the above expression can be rewritten as \((\neg p \vee \neg q)\). Finally, \((\neg p \vee \neg q)\) yields to \((\neg p \downarrow \neg q)\), the expression on the right-hand side. Consequently, our statement \(\neg(p \uparrow q) \Leftrightarrow (\neg p \downarrow \neg q)\) is proven right.

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Most popular questions from this chapter

Use the substitution rules to verify that each of the following is a tautology. (Here \(p, q\), and \(r\) are primitive statements.) a) \([p \vee(q \wedge r)] \vee \neg[p \vee(q \wedge r)]\) b) \([(p \vee q) \rightarrow r] \leftrightarrow[\neg r \rightarrow \neg(p \vee q)]\) c) \([(p \vee q) \rightarrow r] \vee(s \wedge t)] \leftrightarrow[[(p \vee q) \rightarrow r] \vee s] \wedge[(p \vee q) \rightarrow r] \vee t]]\)

a) For primitive statements \(p, q\), find the dual of the statement \((\neg p \wedge-q) \vee\left(T_{0} \wedge p\right) \vee p\). b) Use the laws of logic to show that your result from part (a) is logically equivalent to \(p \wedge \neg q\).

Let \(p(n), q(n)\) represent the open statements $$ p(n): \quad n \text { is odd } \quad q(n): \quad n^{2} \text { is odd } $$ for the universe of all integers. Which of the following statements are logically equivalent to each other? a) If the square of any integer is odd, then the integer is odd. b) \(\forall n[p(n)\) is necessary for \(q(n)]\) c) The square of any odd integer is odd. d) There are some integers whose squares are odd. e) Given any integer whose square is odd, that integer is likewise odd. f) \(\forall n[\neg p(n) \rightarrow \neg q(n)]\) g) Every integer with an odd square is odd. h) Every integer with an even square is even. i) \(\forall n[p(n)\) is sufficient for \(q(n)]\)

Negate and express each of the following statements in smooth English. a) Kelsey will get a good education if she puts her studies before her interest in cheerleading.

For the following statements the universe comprises all nonzero integers. Determine the truth value of each statement. a) \(\exists x \exists y[x y=1]\) b) \(\exists x \forall y[x y=1]\) c) \(\forall x \exists y[x y=1]\) d) \(\forall x \forall y\left[\sin ^{2} x+\cos ^{2} x=\sin ^{2} y+\cos ^{2} y\right]\) e) \(\exists x \exists y[(2 x+y=5) \wedge(x-3 y=-8)]\) f) \(\exists x \exists y[(3 x-y=7) \wedge(2 x+4 y=3)]\)
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