Question 188838
1. p v (q & r)
2. ~r
3. p -> (s -> ~t) 
:. (therefore) ~(s & t) 

Write the conjunction of premises 1 and 2:

       [p v (q & r)] & ~r 

Distribute inside the bracket:

      [(p v q) & (p v r)] & ~r

Use the associative law to move the bracket:

      (p v q) & [(p v r) & ~r]

Use the associative law inside the bracket 
to move the parentheses:

      (p v q) & [(p v (r & ~r)]

(r & ~r) is a contradiction so we replace it by F

      (p v q) & [p v F]

Us the distributive law in revers to "factor" out " p v "

      p v (q v F)

F is the identity for v so we can replace p by p v F

      (p v q) & (p v F)

Use the distributive law to factor out " p v "

      p v (q & F)

Since F is the annihilator for & we can replace (q & F) by F

      p v F

Since F is the identity for v we can replace that by p

       p

Now we take the conjunction of this with premise 3.

       p & [p -> (s -> ~t)]

Since x -> y is equivalent to ~x v y, we
can write the bracket thusly:

       p & [~p v (s -> ~t)]

Use the distributive law:

      (p & ~p) v [p & (s -> ~t)]

(p & ~p) is a contradiction so we can replace it by F

             F v [p & (s -> ~t)]              

Since F is the identity for v, we can eliminate the F v

                  p & (s -> ~t)

Use the fact that x & y -> y, so we can eliminate " p & "

                       s -> ~t

Since x -> y is equivalent to ~x v y, we
can write that as
                       ~s v ~t

We can use DeMorgan's law to "factor" out ~ and get

                      ~(s & t)