# A Friendly Introduction to Mathematical Logic by Christopher C. Leary

By Christopher C. Leary

This undemanding creation to the foremost suggestions of mathematical good judgment makes a speciality of strategies which are utilized by mathematicians in each department of the topic. utilizing an assessible, conversational variety, it techniques the topic mathematically (with distinct statements of theorems and proper proofs), exposing readers to the energy and gear of arithmetic, in addition to its barriers, as they paintings via tough and technical effects. KEY issues: constructions and Languages. Deductions. Comnpleteness and Compactness. Incompleteness--Groundwork. The Incompleteness Theorems. Set concept. : For readers in arithmetic or comparable fields who are looking to know about the most important suggestions and major result of mathematical common sense which are vital to the certainty of arithmetic as a complete.

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Suggestion; You mostly have to worry about uniqueness—for example, suppose that t is c, a constant symbol. How do you know that t is not also a complex term? tn. How do you show that the / and the V s are unique? ] 8. tn for an w-ary relation symbol R) (c) Negation - (d) Disjunction (e) Quantified Also, it must be that if is both = t\t2 and = ^£4, then t\ is identical to tz and £2 is identical to <4, and similarly for other atomic formulas. Furthermore, if (for example) 0 is a negation ( - H I ) , then it must be the case that there is not another formula such that /3), and similarly for disjunctions and quantified formulas.

4. Induction 17 As you look at the proof of this theorem, you notice that there is a base case, when » = 1, and an inductive case. In the inductive step of the proof, we prove the implication If the formula holds for k, then the formula holds for k •+1. We prove this implication by assuming the antecedent, that the theorem holds for a (fixed, but unknown) number k, and from that assumption proving the consequent, that the theorem holds for the next number, k + 1. Notice that this is not the same as assuming the theorem that we are trying to prove.

We have to prove that 211= (Vx3yP(x, y)) [s]. As the statement of interest is universal, we must show that, if c is an arbitrary element of A, 21 |= 3yP(x, y)[s[x|c]], which means that we must produce an element of the universe, d, such that 21 (= P(x, y) [(s[x|c|) [y|