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Let R kbe a -kary relation. Say that Ris definable in Th(,+)if we can give a formulawith kfree variablesx1,...,xksuch that for all,a1,...,ak,(a1,...,ak)is true exactly when .a1,...,akRShow that each of the following relations is definable in.Th(,+)

a.R0={0}

b.R1={1}

c.R=={(a,a)|a}

d.R<={(a,b)|a,banda<b}

Short Answer

Expert verified

a).0(x)=鈭赌y[x+y=y]

b). .data-custom-editor="chemistry" 1(x)=鈭赌y[x+y=y+1]

c).=(x,y)=鈭赌z[0(z)x+z=y]

d).<(x,y)=鈭赌z[0(z)x+z<y]

Step by step solution

01

Solution (a)

In the case R0={0}

There is only one value 0鈥樷赌.

If we add0 in variabley, there will be no effect on y

So the formula: 鈭赌xR0鈭赌y[x+y0+y]

This will become as: 0(x)=鈭赌y[x+y=y]

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