HSS 2016 Solutions to Problem Set V-2

HSS 2016 Solutions to Problem Set V-2 - Deductive Logic Dr...

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Deductive Logic Dr. James Pearson Problem Set V 1. The set { , –} is not expressively adequate. It is trivial that applying the operations of negation and the biconditional repeatedly to a single statement letter will not yield a schema equivalent to “p q”. Let Γ be the set of all compound schemata that can be formed using the connectives in { ≡, –} together with both the statement letters p and q. Consider a schema X drawn from Γ . Then (claim) an even number of T-VAs verify X, and so X is not equivalent to “p q” (which is verified by an odd number of T-VAs). (Indeed, X will not be equivalent to “p . q” or “p q” either.) Proof of claim: Let n be the number of connectives in a schema. Base case: n=1 Let X be a schema from Γ with one connective. Then X is either “p q” or “q p”, which are equivalent, and verified by exactly two T-VAs. Neither is therefore equivalent to “p q”, which is verified by three T-VAs. Inductive step: Let X be a schema from Γ with n>1 connectives. It must have the form R S or –R, where R and S are members of Γ that have strictly less connectives than X. Case: X = –R. By hypothesis, R is verified by an even number of T-VAs. It is thus falsified by an even number of T-VAs. But then, by the fundamental law of negation, an even number of T-VAs must verify –R. X is therefore not equivalent to “p q”, which is verified by three T-VAs. Case: X = R S. By hypothesis, R and S are both verified by an even number of T-VAs.
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