document.write( "Question 1196247: (y→z)&(z→y) (∴(y&z)v(~y&~z)) \n" ); document.write( "

Algebra.Com's Answer #829505 by math_helper(2461) $\"\"$ $\"About$

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document.write( "1. (y-->z) & (z-->y)     Premise
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document.write( "// show (y & z) v (~y & ~z)\r
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document.write( "2.  y-->z       1, Simplification (SIMP)
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document.write( "3.  z-->y       1, SIMP
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document.write( "4.:: y          Conditional Proof (CP) assumption #1
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document.write( "5.:: z          4,2 Modus Ponens (MP)
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document.write( "6.:: y          5,3 MP
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document.write( "// now do the negations
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document.write( "7.:: ~y         CP assumption #2
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document.write( "8.:: ~z         7,3 Modus Tollens (MT)
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document.write( "9.:: ~y         8,2 MT
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document.write( "// at this point we've shown (lines 4-5) y true leads to z true (and 
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document.write( "// lines 5-6, z true leads to y true)
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document.write( "//  and ~y leads to ~z (lines 7-8) and ~z leads to ~y (lines 8-9)
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document.write( "//  In other words, the \"truthness\" of y follows that of z, and vice-versa.
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document.write( "10. (y & z) v (~y & ~z)     4-9, CP\r
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