I have a question relating to Propositional Logic. Any help will be greatly appreciated.

Without changing the meaning of the following formulæ, which rely on operator precedence to be interpreted correctly, introduce brackets in each so that no precedence information is required.

(a) $\lnot p \land q \implies r \land p \land q \land \lnot r \iff F$

Ans $((¬p ∧ q) ⇒ (r ∧p )∧(q ∧¬r ))⇔ F$

(b) $¬p ∧ q ∧ r ⇔ ¬p ∨ ¬q ∧¬r$

Ans $(¬p ∧ (q ∧ r)) ⇔ ¬p ∨ (¬q ∧¬r)$

Please can somebody advise me if my answer is correct or incorrect. Thank you all geniuses so much.


You're almost there. The missing parens are inserted & highlighted:

(a) $((\lnot p \land q) \rightarrow \color{red}{\bf(}(r \land p) \land (q \land \lnot r)\color{red}{\bf)}) \leftrightarrow F$

(b) $(\lnot p \land (q \land r)) \leftrightarrow \color{red}{\bf(}\lnot p \lor (\lnot q \land \lnot r)\color{red}{\bf)}$

But since they want no precedence information assumed, we have to make the formulas uglier:

(a) $((\color{red}{\bf(}\lnot p\color{red}{\bf)} \land q) \rightarrow ((r \land p) \land (q \land \color{red}{\bf(}\lnot r\color{red}{\bf)}))) \leftrightarrow F$

(b) $(\color{red}{\bf(}\lnot p\color{red}{\bf)} \land (q \land r)) \leftrightarrow (\color{red}{\bf(}\lnot p\color{red}{\bf)} \lor (\color{red}{\bf(}\lnot q\color{red}{\bf)} \land \color{red}{\bf(}\lnot r\color{red}{\bf)}))$

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  • $\begingroup$ I don’t think that (a) is quite right: surely the expression between the red parentheses ought to be $$\big((r\land p)\land q\big)\land(\neg r)\;.$$ $\endgroup$ – Brian M. Scott Oct 27 '13 at 7:13
  • $\begingroup$ @BrianM.Scott That would be more canonical, but this does meet the requirements of "no operator precedence information required" and "without changing the meaning." $\endgroup$ – Trevor Wilson Oct 27 '13 at 7:16
  • $\begingroup$ @Trevor: Not quite: it still requires parentheses around $\neg r$. $\endgroup$ – Brian M. Scott Oct 27 '13 at 7:17
  • $\begingroup$ @Brian Oh, good point. I didn't notice that part. $\endgroup$ – Trevor Wilson Oct 27 '13 at 7:17
  • 1
    $\begingroup$ @Trevor: I’m guilty too: I didn’t notice until just now that the same applies elsewhere. $\endgroup$ – Brian M. Scott Oct 27 '13 at 7:19

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