document.write( "Question 1171792: 1. A v B
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Algebra.Com's Answer #796787 by math_helper(2461) $\"\"$ $\"About$

You can put this solution on YOUR website!


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document.write( "1. A v B            Premise
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document.write( "2. A <--> (C & D)   Premise
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document.write( "3. B --> (D & G)    Premise
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document.write( "::4. A              Conditional Proof (CP) assumption #1
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document.write( "::5. (A --> (C & D)) & ((C & D) --> A)   2 Biconditional Equivalence (not sure of proper term)    
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document.write( "::6. (A --> (C & D))     5  Simplification (SIMP)
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document.write( "::7.  C & D              6,4 Modus Ponens (MP)
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document.write( "::8. D                   7  SIMP
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document.write( "::9. A --> D             4-8, CP 
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document.write( "::9. B                   CP assumption #2
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document.write( "::10. D & G              3,9  MP
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document.write( "::11. D                  10  SIMP
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document.write( "::12. B --> D            9-11 CP
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document.write( "::13. (A v B) --> D      4-12 CP Proof By Cases (PBC)
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document.write( "14. (A v B) --> D        4-13 CP (discharges CP assumptions)
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document.write( "15. D                    14,1  MP      \r
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document.write( "What does this proof do?  Lines 4-8 show if we assume A true, D is true (if A then D).  Lines 9-13 show \"if B then D\", therefore \"if (A v B) then D\" is true because we've covered the two possible cases (A true or B true) and we promote that result on line 14.  Line 15 follows from the premise on line 1.
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