Introduction to Logic: Problems and solutions
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Solution Hints to the Exercises A Concise Introduction to
A Concise Introduction to. Mathematical Logic by Wolfgang Rautenberg The reader may mail improved solutions to the author whose website is.
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Introduction to Logic: Problems and solutions
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In this section we look at how philosophers attempt to answer such questions in a systematic and rational way. This section also introduces the fields of
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Solution Hints to the Exercises
fromA Concise Introduction to
Mathematical Logic
by Wolfgang RautenbergThird Edition, Springer Science
+Business MediaLast version September 2010
Hints to the exercises can as a rule easily be supplemented to a complete solution. Exercises that are essential for the text are solved completely. The reader may mail improved solutions to the author whose website is www.math.fu-berlin.de/~raut. 12 Solution hints to the Exercises
Section 1.1
1. (a): xkis fictional infiffak= 0. (b): Because of the uniqueness, 2 n+1(= number of subsets off0;:::;ng) is the number sought for. (c): induction on formulas in:;+andp1;:::;pn. 2. Consider on Fthe propertyE': ''is prime or there are;2Fwith '=:or'= ()where=^or=_." Formula induction showsE'for all'2F. 3. V erifyb yinduction on 'the propertyE': 'no proper initial segment of'is a formula nor can'be a proper initial segment of a formula". Induction step: Case'=:. Then a proper initial segment of: either equals:(hence is not a formula), or has the form:whereis a proper initial segment of. Thus =2Fby the induction hypotheses, hence also: =2F(since a formula starting with:must have the form :for some formulaby Exercise 2). Case'= (). Letbe a proper initial segment of'or conversely. Assume thatis a formula so that= (000), some0;02F(Exercise 2). Then6=0, for otherwise necessarily='. Hence0is a proper initial segment of or conversely, a contradiction to the induction hypothesisE. 4.Assume that () = (000), hence=000. If6=0then
is a proper initial segment of0or conversely. This is impossible byExercise 3. Consequently=0, hence=0and=0.
Section 1.2
1.w((p!q1)^(:p!q2)) = 0iffwp= 1;wq1= 0orwp= 0;wq2= 0,
and the same condition holds forw(p^q1_:p^q2) = 0. In a similar way the second equivalence is treated.2.:pp+ 1,1p+:p,p$qp+:q,p+qp$ :q :(p$q).
3. Induction on the 2Fnf0;1;^;_g(= set of formulas in0;1;^;_and p1;:::pn). Iff;g2Bnare monotonic then so is~a7!f~ag~a, whereis
^or_. For simplicity, treat first the casen= 1. Converse: Induction on the arityn. Clear forn= 0, with the formulas 0 and 1 representing the two constants. Withf2Bn+1alsofk:~x7!f(~x;k)is monotonic (k= 0;1). Letk2Fnf0;1;^;_grepresentfk(induction hypothesis). Then0_(1^pn+1)representsf. Note thatw06w1for allw.Solution hints to the Exercises 3
4. By Exercise 3, a not repr esentablef2Bn+1is not monotonic in the last argument, say. Thenf(~a;1) = 0andf(~a;0) = 1for some ~a2 f0;1gn, henceg:x7!f(~a;x)is negation. This proves the claim.Section 1.3
1. (a): MP easily yiel dsp!q!r,p!q,pr. Apply (D) three times. 2.The deduction the oremyields (!)!(
3.Assume that wX;_. Then clearlywX;orwX;.
5.Let X` =2XThenX;`for each. Thus,X`by (T).
Section 1.4
1.X[f:j2Yg `?)X[f:0;:::;:ng `?, some0;:::;n2Y.
HenceX`(V
i6n:i)!?, or equivalently,X`W i6nai. This all is easily verified if`is replaced by. 2.Supplemen tLem ma4.4 b ypro vingX`_,X`orX`.
3.Cho oseX;'such thatX0'andX`0'. LetYX[ f:'gbe
maximally consistent in`. Definebyp=>forp2Yandp=:> otherwise. Induction onyields with the aid of(^)and(:)page 28 ()2Y) `; =2Y) ` :.In proving(),`>,`) ` ::,:` :(^), and:` :(^)
are needed which easily follow from the:rules. By(),` :', hence0:'. Clearly`Y(i.e.,`for all2Y), and so`0Y. But
Y `0'(substitution invariance). Thus,`0'. Therefore`0for all by(:1), so that`is maximal by definition. 4. There is a smallest co nsequencerelation with the prop erties(^1)-(:2), namely the calculus`of this section. Since` and`is already maximal according to Exercise 3,`andmust coincide.Section 1.5
1. F orfinite Measily shown by induction on the number of elements of M. Note thatMhas a maximal element. General case: Add to the formulas in Example 1 the set of formulasfpabja60bg.4 Solution hints to the Exercises
2.):AssumeM;N =2F. Thenn(M[N) =nM\nN2F, because
:M;:N2F. ThereforeM[N =2F.(:M2FimpliesM[N2F by condition (b). For proving(:)from(\)observe thatM[nM2F.3.): LetUbe trivial, i.e.,E2Ufor some finiteEI. Induction on
the number of elements inEand Exercise 2 easily show thatfi0g 2U for somei02E. The converse is obvious.Section 1.6
1. First v erifyth ededucti ontheore m,whic hh oldsfor eac hcalculus with MP as the only rule and A1, A2 among the axioms; cf. Lemma 6.3.Xis consistent iffX0?, forX`?)X`(!?)!?=::by
A1, henceX`by A3. Now proveX`!iffX`)X`,
providedXis maximally consistent. This allows one to proceed along the lines of Lemma 4.5 and Theorem 4.6. 2. Apply Zorn"s lem mato H:=fYXjY0g. Note that ifKHis a chain thenSK2Hdue to the finitarity of`. 3. (a): Suc ha set Xsatisfies():X`'!for all. For otherwise X;' !`', henceX`('!)!', and soX`'by Peirce"s axiom. Suppose =2X. ThenX;`';'!by(), and soX;`. (b): With (a) easily followsX`!iffX`)X`as in Exercise 1. Proceed with an adaptation of Lemma 4.5. 4.Crucial for completeness is the pro ofof (m): `)
by induction on the rules of`. (m) implies (M):X;`)X; proving first that a calculus`based solely on unary rules satisfiesX`)`for some2X. E.g.,`)
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