Question 1196247
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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