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## Foundations of Mathematics I

by: Elaina Osinski

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# Foundations of Mathematics I MATH 558

Marketplace > Pennsylvania State University > Mathematics (M) > MATH 558 > Foundations of Mathematics I
Elaina Osinski
Penn State
GPA 3.88

Simpson

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Simpson
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## Popular in Mathematics (M)

This 0 page Class Notes was uploaded by Elaina Osinski on Sunday November 1, 2015. The Class Notes belongs to MATH 558 at Pennsylvania State University taught by Simpson in Fall. Since its upload, it has received 20 views. For similar materials see /class/232997/math-558-pennsylvania-state-university in Mathematics (M) at Pennsylvania State University.

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Date Created: 11/01/15
A Semigroup With Unsolvable Word Problem Stephen G Simpson Original Draft February 10 2000 Current Draft February 8 2001 We construct a nitely presented semigroup with unsolvable word probleml We follow the exposition of the first few sections of Chapter 12 of Joseph J Rot man The Theory of Groups 2nd edition AllynBacon 1973 with the difference that Rotman uses Turing machines while we use register machinesl Our construction is based on the following fact There is a register machine program 73 which computes the partial recursive function 2 gt gt 0 0551 I and such that 73 uses only two registers R1 and R2 This follows easily from Exercise 59 in my Math 558 lecture notes Foundations of Mathematicsl Note that I 2 731 halts is nonrecursivel In other words given I the problem of deciding whether 73 halts if started with z in R1 and with R2 empty is unsolvablel Furthermore we may safely assume that if 731 halts then it halts with both registers empty The idea of our construction is to encode the action of 73 into the word problem of a semigroup 5 Let 1 l l l 1 be the instructions of 73 As usual 11 is the first instruction executed and Io is the halt instructionl Our semigroup S will have 1 3 generators ab qoq1 l qll If R1 and R2 contain 1 and y respectively and if m is about to be executed then we represent this state as a word axqmay Thus a serves as a counting token and 12 serves as an endofcount markerl For each m 1H if m says increment R1 and go to Inc we represent this as a production qm A aqnO or as a relation qm aqnol If m says increment R2 and go to Inc we represent this as a production qm A qnoa or as a relation qm qnoal If m says ile is empty go to Inc otherwise decrement R1 and go to 1m we represent this as a pair of productions qu A bqno aqm A gm or as a pair of relations 12an bqno aqm qnll If m says if R2 is empty go to no otherwise decrement R2 and go to 1m we represent this as a pair of productions qmb A qnob qma A qm or as a pair of relations qmb qnob qma qml Thus the total number of productions or relations is 1 211 where l 1 l and 1 is the number of increment instructions and l is the number of decrement instructionsl Let S be the semigroup described by these generators and relations We claim that for all z baqub bqob in S if and only if 731 haltsl The if77 part is clear For the only if 7 part assume that baqub 124012 in S This implies that there is a sequence of words baqub we wn bqob where each wi1 is obtained from wi by a forward or backward production We claim that the backward productions can be eliminated In other words if there are any backward productions we can replace the sequence we wn by a shorter sequence This is actually obVious because if there is a backward production then there must be one which is immediately followed by a forward production and these two must be inverses of each other because 73 is deterministici Thus we see that baqub 2401 Via a sequence of forward productionsi This implies that 731 haltsi Our claim is proved

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