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I would like to reduce this statement:$$ P \veebar (P \implies B) $$ using only $\neg$, $\land$ I've found this solution but I don't know if I'm wrong: $$\neg(P \land B)$$ Because the book proposes this one using XOR: $$\neg P \veebar (P \land \neg B)$$ Is there any errors or things by which I should do the same?


Exercise 1.8-4 p.16 from Analisi matematica vol.1 by Enrico Giusti

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  • $\begingroup$ What your book suggests is only a hint, not a solution. The one you have arrived at is correct, see WolframAlpha. $\endgroup$ – dtldarek May 8 '14 at 11:39
  • $\begingroup$ @dtldarek thanks for the link! in fact is with all the others solutions at the end of the chapters, maybe the author miss this solution... $\endgroup$ – G M May 8 '14 at 11:44
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That's right. To check, let's unpack the initial formula:

$$P \oplus (P \rightarrow B)$$

$$P \oplus (\lnot P \lor B)$$

$$(P \lor (\lnot P \lor B))\land \lnot(P \land (\lnot P \lor B))$$

$$(\top \lor B)\land \lnot(P \land (\lnot P \lor B))$$

$$\top\land \lnot(P \land (\lnot P \lor B))$$

$$\lnot(P \land (\lnot P \lor B))$$

$$(\lnot P \lor \lnot(\lnot P \lor B))$$

$$(\lnot P \lor (P \land \lnot B))$$

$$(\lnot P \lor P) \land (\lnot P \lor \lnot B)$$

$$\top \land (\lnot P \lor \lnot B)$$

$$(\lnot P \lor \lnot B)$$

Which, once De Morganed, is equivalent to your solution:

$$\lnot(P \land B)$$

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Writing out the truth tables for both we obtain the following:

P   B  (P⟹B)  [P⊻(P⟹B)]  ¬(P∧B)
0   0     1       1         1
0   1     1       1         1
1   0     0       1         1
1   1     1       0         0

Since all rows match here, it follows that your solution is correct.

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