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Please, help me make equivalent transformations with this formula (A∨C→B)(A→C)(¬B→¬A∧C)(¬A→(C→B))(B→¬C→¬A). Thanks.

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    $\begingroup$ This doesn't make sense. $\endgroup$ – Git Gud Jun 4 '14 at 21:59
  • $\begingroup$ I must make as here, only with my formula minus.com/i/s3dtb4Vh0FCL $\endgroup$ – Maxikkk Jun 5 '14 at 8:30
  • $\begingroup$ oh... I'm stupid. Sorry, of course (A+C→B)(A→C)(¬B→¬A∧C)(¬A→(C→B))(B→¬C→¬A). Thank you. $\endgroup$ – Maxikkk Jun 5 '14 at 9:10
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You cannot "mix" in this way different "conventions" regarding symbols.

If you want to use propositional connectives (like : $\lor, \land$) instead of boolean operators (like : $\cdot, +$) you have to rewrite your formula with $\lor$ in place of $+$ and $\land$ in place of $\cdot$ (justaxposition).

If so, I think that your formula must be :

$((A∨C)→B) \land (A→C) \land (¬B→(¬A \land C)) \land (¬A→(C→B)) \land (B→(¬C→¬A))$

having restored some missing parentheses.

Then we eliminate $\rightarrow$, through the equivalence between $P \rightarrow Q$ and $\lnot P \lor Q$.

In this way, splitting the problem, we have five conjuncts to consider:

(i) $((A∨C)→B)$ is : $\lnot (A∨C) \lor B$ which, by De Morgan, is : $(\lnot A \land \lnot C) \lor B$, which in turn is equivalent to : $(\lnot A \lor B) \land (B \lor \lnot C)$, by distributivity.

In boolean form is : $(\bar {A} + B)(B + \bar {C})$.

(ii) $(A→C)$ is simply : $(\lnot A \lor C)$. In boolean : $(\bar {A} + C)$.

(iii) $(¬B→(¬A \land C))$ is : $B \lor (\lnot A \land C)$, using double negation, which in turn is equivalent to : $(\lnot A \lor B) \land (B \lor C)$, by distributivity.

In boolean form is : $(\bar {A} + B)(B + C)$.

(iv) $(¬A→(C→B))$ is : $A \lor (\lnot C \lor B)$. In boolean : $(A + B + \bar {C})$.

(v) $(B→(¬C→¬A))$ is : $\lnot B \lor (C \lor \lnot A)$. In boolean : $(\bar {A} + \bar {B} + C)$.

Now, we can "reassemble" the conjuncts (i) to (v) without redundant terms :

$(\bar {A} + B)(B + \bar {C})(\bar {A} + C)(B + C)(A + B + \bar {C})(\bar {A} + \bar {B} + C)$.

Now we start with boolean simplification, "inserting" the missing terms [i.e. : $(\bar {A} + B)$ is rewritten as : $(\bar {A} + B + C\bar {C})$ i.e. $(\bar {A} + B + C)(\bar {A} + B + \bar {C})$ ] :

$(\bar {A} + B + C)(\bar {A} + B + \bar {C})(A + B + \bar {C})(\bar{A} + B + \bar {C})(\bar {A} + B + C)(\bar {A} + \bar {B} + C)(A + B + C)(\bar {A} + B + C)(A + B + \bar {C})(\bar {A} + \bar {B} + C)$

and cancel the redundant terms :

$(\bar {A} + B + C)(\bar {A} + B + \bar {C})(A + B + \bar {C})(\bar {A} + \bar {B} + C)(A + B + C)$.

This is the standard Product-of-Sums (POS) Form.

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