Ummm - keep simplifying using

a -> b = !a + b

!(a + b) = !a * !b

!(ab) = !a + !b

where ! means "not", + means "or", and * or multiplication means "and".

To prove that

{[( not p v q) --> r] and [r --> (s v t)] and (not s and not u) and ( not u --> not t)} -->p

= ((!p + q) -> r)(r -> (s+t))(!s)(!u)(!u -> !t)) -> p

= (p(!q)+r)(!r+s+t)(!s)(!u)(u+!t)) -> p

Now (!u)(u+!t) = !u !t

(!r+s+t)(!s)(!t) = !r !s !t

So LHS = (p !q + r)(!r)(!s)(!t)(!u) =

p(!q)(!r)(!s)(!t)(!u)

SO

To prove that

p(!q)(!r)(!s)(!t)(!u) -> p or

!p + q + r + s + t + u + p

which is true since it has !p + p