![]() ![]() Thus any wff of length 2 or 3 must not contain any connectives. By lemma 1 (from this document), any wff with connectives will be of length at least 4. All wffs must have at least one sentence symbol. Show that there are no wffs of length 2, 3, or 6, but that any other postive length is possible. By a similar argument, the number of left parenthesis, right parenthesis and connective symbols in (α ∧ β), (α ∨ β), (α → β) and (α ↔ β) are equal. lγ = lα + 3, rγ = rα + 3, cγ = cα + 3 = so lγ = rγ = cγ by the induction hypothesis. Induction hypothesis: Let α and β be two wffs. Let l, r and c be, respectively, the number of left parenthesis, right parenthesis and connective symbols in a wff. Proof: We will use the induction principle. 1 Sentential Logic 1.1 The Language of Sentential Logic Lemma 1 The number of left parenthesis, right parenthesis and connective symbols are equal in a wff. Do feel free to email me if you have any comments or to point out any errors (of which there will likely be plenty). It’d be great if this document helps anyone out in any form or fashion. I have posted it online as a possible resource for others also using the book. Its main purpose is to facilitate my own learning. ![]() Some Solutions to Enderton’s Mathematical Introduction to Logic Kelvin Soh ∗ OctoOverview This document details my attempt to solve some of the problems in Herbert Enderton’s A Mathematical Introduction to Logic (2nd Edition). ![]()
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