Prove the argument:

1.  p -> q
2.  r \/ s 
3.  ~s -> ~t
4.  ~q \/ s
5.  ~s
6.  (~p \/ r) -> u
7.  w \/ t
therefore u /\ w


Change the disjunctions in 2, 4, and 7, to their equivalent conditionals 
by this rule  a \/ b <=> ~a -> b

1.  p -> q
2.  ~r -> s 
3.  ~s -> ~t
4.  q -> s
5.  ~s
6.  (~p \/ r) -> u
7.  ~w -> t

Now write all the equivalent contrapositives of the conditionals:

 8.  ~q -> p
 9.  ~s -> r 
10.    t -> s
11.  ~s -> ~q
12.  ~s
13.  ~u -> ~(~p \/ r) 
14.  ~t -> w


Start with 5

      ~s 

By 9, we have

      ~s -> r

By the conditional r -> (r \/ ~p), we have

      ~s -> r -> (r \/ ~p)

By the commutative law, (r \/ ~p) <=> (~p \/ r), we have:

      ~s -> r -> (r \/ ~p) <=> (~p \/ r)

By 6, we have 

      ~s -> r -> (r \/ ~p) <=> (~p \/ r) -> u

That is one part of the conjunction conclusion

Start again with 5

      ~s

By 3, we have

      ~s -> ~t

By 14, we have

      ~s -> ~t -> w

So by syllogism we have both parts
of the conjunction conclusion u and w.

Therefore u /\ w

q was not involved so we did not need 1, 4,
or their contrapositives 8, 11.

Edwin