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Back to the initial problem:
Back to the initial problem:
* Show that <math>\lnot (p \Leftrightarrow q)</math> is equivalent to <math>(\lnot q) \Leftrightarrow p.</math>
:* Show that <math>\lnot (p \Leftrightarrow q)</math> is equivalent to <math>(\lnot q) \Leftrightarrow p.</math>
We can translate this into logical graphs by supposing that we have to express everything in terms of negation and conjunction, using parentheses for negation and simple concatenation for conjunction. In this way of assigning logical meaning to graphical forms — for historical reasons called the "existential interpretation" of logical graphs — basic logical operations are given the following expressions:
We can translate this into logical graphs by supposing that we have to express everything in terms of negation and conjunction, using parentheses for negation and simple concatenation for conjunction. In this way of assigning logical meaning to graphical forms — for historical reasons called the "existential interpretation" of logical graphs — basic logical operations are given the following expressions:
:* [[Logical_graph#C3._Dominant_form_theorem|C<sub>3</sub>. Dominant Form Theorem]]
:* [[Logical_graph#C3._Dominant_form_theorem|C<sub>3</sub>. Dominant Form Theorem]]
In my experience with a variety of propositional calculi, the logical graph picture is almost always the best way to see ''why'' a theorem is true. In the example at hand, most of the work was already done by the time we wrote down the problem in logical graph form. All that remained was to see the application of the generation and double negation theorems to the left and right sides of the equation, respectively.
In my experience with a number of different propositional calculi, the logical graph picture is almost always the best way to see ''why'' a theorem is true. In the example at hand, most of the work was already done by the time we wrote down the problem in logical graph form. All that remained was to see the application of the generation and double negation theorems to the left and right sides of the equation, respectively.