SOLUTION: (y→z)&(z→y) (∴(y&z)v(~y&~z))

1. (y-->z) & (z-->y)     Premise

// show (y & z) v (~y & ~z)



2.  y-->z       1, Simplification (SIMP)

3.  z-->y       1, SIMP

4.:: y          Conditional Proof (CP) assumption #1

5.:: z          4,2 Modus Ponens (MP)

6.:: y          5,3 MP

// now do the negations

7.:: ~y         CP assumption #2

8.:: ~z         7,3 Modus Tollens (MT)

9.:: ~y         8,2 MT

// at this point we've shown (lines 4-5) y true leads to z true (and 

// lines 5-6, z true leads to y true)

//  and ~y leads to ~z (lines 7-8) and ~z leads to ~y (lines 8-9)

//  In other words, the "truthness" of y follows that of z, and vice-versa.

10. (y & z) v (~y & ~z)     4-9, CP