Proving (s ^ ~p) ==> t using Sentential Calculus

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To prove (s ^ ~p) ==> t using sentential calculus, start with the premises ~(q ^ s) and q OR p. From the first premise, derive (~q OR ~s) using De Morgan's laws. Then, combine this with the second premise to obtain p OR ~s. To show the desired implication, consider the contrapositive of the hypothesis, which can simplify the proof process. This approach effectively leads to the conclusion that (s ^ ~p) implies t.
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Question:

Using sentential calculus (with a four column format), prove that the conclusion (s ^ ~p) ==> t follows from the premises: ~(q ^ s) and q OR p. (Hint: Employ conditionalization).

Work done:
Code:
   (1)  ~(q ^ s)         P
   (2)  (~q OR ~s)     DeM (1)
   (3)  q OR p            P
   (4)  p OR ~s          Cut (2, 3)
   (5)  ~(s ^ ~p)       DeM (4)
but, of course I'm to show (s ^ ~p) ==> t.

How should I go about it? thank you.
 
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contrapositive of the hypothesis seems to work! :-)
 

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