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I am currently trying to nail down tautologies. However I tried solving a question and got a bit confused its asking for solving by logical equivalences.

This are the questions:

$$\begin{array}{c} \neg(\neg p\wedge p)\vee q \\ \underbrace{\begin{array}{c|c|c|c|c} p&q&(\neg p\wedge p)&\neg(\neg p\wedge p)&\neg(\neg p\wedge p)\vee q \\\hline T&F&F&T&T \\ F&T&T&F&T \\ T&T&F&T&T \\ F&F&F&T&T \\ \end{array}}_{\text{Tautology}} \\ \\ (p\Rightarrow p)\wedge\neg(q\Leftrightarrow q) \\ \underbrace{\begin{array}{c|c|c|c|c|c} p&q&(p\Rightarrow p)&(q\Leftrightarrow q)&\neg(q\Leftrightarrow q)&(p\Rightarrow p)\wedge\neg(q\Leftrightarrow q) \\\hline T&F&T&T&F&F \\ F&T&T&T&F&F \\ T&T&T&T&F&F \\ F&F&T&T&F&F \\ \end{array}}_{\text{Contradiction}} \end{array} $$

Thank you

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  • $\begingroup$ Is there a typo in the third column first table? Not p and p? $\endgroup$ – jdods May 17 '18 at 3:21
  • $\begingroup$ I think you entered your elements incorrectly. Consider $\neg (\neg p \wedge p)$. What is its value? $\endgroup$ – David G. Stork May 17 '18 at 3:21
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Have a look at the final column in both your examples. In the final column of the first one, all are true, so it's a tautology. In the final column of the second one, they're all false, so it's a contradiction.

I'll need more context to say anything about your statement about logical equivalences.

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Well, not quite so.   The column for $\neg p\wedge p$ should be false in every row; making $\neg (\neg p\wedge p)$ true in every row .

Otherwise the truth tables are okay.

However, they are redundant except as a matter of exercise.   You don't need them to solve this.


Since $\neg p\wedge p$ is a contradiction, clearly ${\neg(\neg p\wedge p)\vee q}$ will be a tautology.

Since $q\leftrightarrow q$ is a tautology, clearly ${(p\to p)\wedge\neg (q\leftrightarrow q)}$ must be a contradiction.

Think about the reason why that might be so.

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