Let the operations *, odot in {wedge, vee}. I | vedclass

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(B) We check each option by evaluating the expression $(p * q) \odot (p \odot \sim q)$.For option $A$: $* = \vee, \odot = \wedge$.The expression becom

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$(\vee, \vee)$

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(B) We check each option by evaluating the expression $(p * q) \odot (p \odot \sim q)$.For option $A$: $* = \vee, \odot = \wedge$.The expression becomes $(p \vee q) \wedge (p \wedge \sim q)$.Using the distributive law: $p \vee (q \wedge \sim q) \wedge (p \wedge \sim q)$ is not correct. Let's re-evaluate: $(p \vee q) \wedge (p \wedge \sim q) \equiv (p \wedge p \wedge \sim q) \vee (q \wedge p \wedge \sim q) \equiv (p \wedge \sim q) \vee (F) \equiv p \wedge \sim q$,which is not a tautology.For option $B$: $* = \vee, \odot = \vee$.The expression becomes $(p \vee q) \vee (p \vee \sim q)$.By associativity and commutativity: $p \vee p \vee (q \vee \sim q) \equiv p \vee T \equiv T$.Since the result is $T$ (a tautology),option $B$ is correct.

Let the operations $, \odot \in \{\wedge, \vee\}$. If $(p q) \odot (p \odot \sim q)$ is a tautology,then the ordered pair $(*, \odot)$ is:

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Let the operations $*, \odot \in \{\wedge, \vee\}$. If $(p * q) \odot (p \odot \sim q)$ is a tautology,then the ordered pair $(*, \odot)$ is: