Question 178748
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(1)  ~C <font face = "arial">\/</font> D 
(2)  (~A <font face = "arial">\/</font> B) -> F 
(3)  ~B -> C 
(4)  ~F 
Prove: D

   I.  ~F                          Given 

  II.  ~(~A<font face = "arial">/\</font>~B)                   By ~F -> ~(~A<font face = "arial">\/</font>B), the contrapositive of (2)

 III.  ~(~A)<font face = "arial">/\</font>~B                   ~(~A<font face = "arial">\/</font>B) <-> ~(~A)<font face = "arial">/\</font>~B DeMorgan's law

  IV.  A<font face = "arial">/\</font>~B                       ~(~A)<-> A  Negation of negation 

   V.  ~B                          Conjunction IV                   

  VI.  C                           V, and (3),  ~B -> C is given

 VII.  ~C <font face = "arial">\/</font> D                     given

VIII.  C <font face = "arial">/\</font> (~C <font face = "arial">\/</font> D)              Conjunction of VI with VII
 
  IX.  (C <font face = "arial">/\</font> ~C) <font face = "arial">\/</font> (C <font face = "arial">/\</font> D)       Distributive law with VIII

   X.  false <font face = "arial">\/</font> (C <font face = "arial">/\</font> D)           Conjunction of C with its negation is false 

  XI.  C <font face = "arial">/\</font> D                     "false" is the identity for disjunction       

 XII.  D                           Conjunction XI

Edwin</pre>