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My original excercise is, to use the truthtable, but do i have to say anything here?

$\begin{array}{ccc|c@{}ccc@{}ccc@{}c@{}ccc@{}c@{}ccc@{}ccc@{}c} A&B&C&(&A&\Rightarrow&(&B&\Rightarrow&C&)&)&\Leftrightarrow&(&(&A&\land&B&)&\Rightarrow&C&)\\\hline 1&1&1&&1&1&&1&1&1&&&\mathbf{1}&&&1&1&1&&1&1&\\ 1&1&0&&1&0&&1&0&0&&&\mathbf{1}&&&1&1&1&&0&0&\\ 1&0&1&&1&1&&0&1&1&&&\mathbf{1}&&&1&0&0&&1&1&\\ 1&0&0&&1&1&&0&1&0&&&\mathbf{1}&&&1&0&0&&1&0&\\ 0&1&1&&0&1&&1&1&1&&&\mathbf{1}&&&0&0&1&&1&1&\\ 0&1&0&&0&1&&1&0&0&&&\mathbf{1}&&&0&0&1&&1&0&\\ 0&0&1&&0&1&&0&1&1&&&\mathbf{1}&&&0&0&0&&1&1&\\ 0&0&0&&0&1&&0&1&0&&&\mathbf{1}&&&0&0&0&&1&0& \end{array}$

To me that's self explanatory. Do i have to write anything here?

And is it possible to do it without a truthtable?

Note: This is a homework, so please no full solutions. Anyways I'm more interested in how to do it rather then how it is as a result.

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  • $\begingroup$ You can do this using rules of Boolean algebra. You need not write anything else. $\endgroup$ Commented Oct 23, 2016 at 8:04

1 Answer 1

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Yes, it is possible to do this without truth table.

$A \Rightarrow B$ means either $A$ is false (or) $B$ is true. So, $(A \Rightarrow B)$ $\Leftrightarrow \overline{A} \lor B$

$A \Rightarrow (B \Rightarrow C) \Leftrightarrow (A \Rightarrow (\overline{B} \lor C)) \Leftrightarrow \overline{A} \lor \overline{B} \lor C$

$\color{olive}{(A \wedge B) \Rightarrow C \Leftrightarrow (A \land B \Rightarrow C) \Leftrightarrow \overline{AB} \lor C \Leftrightarrow \overline{A} \lor \overline{B} \lor C}$

Thus, both LHS and RHS are equivalent

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