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The proposition $(P \vee Q) \vee (\sim P \wedge \sim Q)$ is a:

(A) contradiction (B) tautology (C) neither a contradiction nor a tautology (D) both a contradiction nor a tautology
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Let us construct the truth table for this proposition:
$ \begin{matrix} P & Q & P\vee Q & \sim P & \sim Q & \sim P \wedge \sim Q & (P \vee Q) \vee (\sim P\; \wedge \sim Q)\\ T & T &T & F & F & F &T \\ F& T & T &T & F &F & T\\ T&F & T & F & T&F &T \\ F& F & F & T& T &T &T \end{matrix}$
Since the proposition is True for every assignment of truth values to its components, it is a tautology.
answered Mar 20, 2014 by balaji.thirumalai

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