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

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Question 1196247: (y→z)&(z→y) (∴(y&z)v(~y&~z))
Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!

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

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