Changes

===Problem===

===Problem===

[http://mathforum.org/kb/message.jspa?messageID=6513648 Problem posted by Mike1234 on the Discrete Math List at the Math Forum].

[http://mathforum.org/kb/message.jspa?messageID=6513648 Problem posted by Mike1234 on the Discrete Math List at the Math Forum ].

* Required to show that <math>\lnot (p \Leftrightarrow q)</math> is equivalent to <math>(\lnot q) \Leftrightarrow p.</math>

* Required to show that <math>\lnot (p \Leftrightarrow q)</math> is equivalent to <math>(\lnot q) \Leftrightarrow p.</math>

===Solution===

===Solution===

[http://mathforum.org/kb/plaintext.jspa?messageID=6514666 Solution posted by Jon Awbrey, using the calculus of logical graphs].

[http://mathforum.org/kb/plaintext.jspa?messageID=6514666 Solution posted by Jon Awbrey, working by way of logical graphs].

In logical graphs, the required equivalence looks like this:

In logical graphs, the required equivalence looks like this: