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

[lizzyb]

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:

   (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.

 

[lizzyb]

contrapositive of the hypothesis seems to work! :-)

 

[piercas]

To prove (s ^ ~p) ==> t, we can use conditionalization, which states that if we have a conditional statement, we can derive the consequent (t) from the antecedent (s ^ ~p) by assuming the antecedent and then proving the consequent. In this case, we can assume (s ^ ~p) and then prove t.

(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)
(6) s ^ ~p Assumption
(7) s Simplification (6)
(8) ~p Simplification (6)
(9) q Disjunctive syllogism (3, 7)
(10) t Modus ponens (9, 5)
(11) (s ^ ~p) ==> t Conditionalization (6-10)

Therefore, we have proven that (s ^ ~p) ==> t using sentential calculus.