By Norman Sullivan
Discover the place you rank at the IQ scale via taking this interesting, self-scoring attempt.
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Extra info for Test Your Own IQ Again
Hence, |= (A1 ∧ · · · ∧ An ) → (B ∧ (¬B)) by the Deduction theorem. But for any assignment to atoms in Ai and B, v(B ∧ (¬B)) = F. Now let A1 , . . , An be satisfiable. Hence , there is an assigmment v such that v(Ai ) = T, i = 1, . . , n. Extend this assignment to v ′ so that it is an assignment for any different atoms that might appear in B. Thus, v ′ (A1 ∧ · · · ∧ An ) → (B ∧ (¬B)) = F. This contradicts the stated Deduction theorem. Hence, inconsistency implies not satisfiable. Conversely, suppose that A1 , .
We must now use the original MP steps to continue the new demonstration construction since A might be in the original demonstration an MP step or the last step of the original demonstration might be an MP step. After using all of the original non-MP steps, we now renumber, for reference, all the original MP steps. We now define by induction on the number m of MP steps a procedure which will complete our proof. Case (m = 1) We have that there is only one application of MP. Let Bj be this original MP step.
2 Determine whether or not the set of premises A ↔ B, B → C, (¬C) ∨ D, (¬A) → D, ¬D. (written in formula variable form) is consistent. First, re-express this set in terms of atoms. P ↔ Q, Q → R, (¬R) ∨ S, (¬P ) → S, ¬S. (a) Let v(¬S) = T ⇒ S = F. (b) Let v((¬R) ∨ S) = T ⇒ v(R) = F. (c) Let v(Q → R) = T ⇒ v(Q) = F. (d) Let v(P ↔ Q) = T ⇒ v(P ) = F ⇒ v((¬P ) → S) = F. Thus the set is inconsistent and, hence, we can substitute the original formula variables back into the set of premises. If |= is associated with ordinary propositional deduction, then it should also mirror the propositional metalogic we are using.
Test Your Own IQ Again by Norman Sullivan