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EvanOktavianus

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I am reading Suppes' book on axiomatic set theory and having difficulties understanding the part on formal conditional definition.

**Background**

in p.18, he gave the rule for operator conditional definition as follows:

An implication P introducing a new operation symbol O is a conditional definition iif P is of the form:

(1) Q → [O(v[itex]_{1}[/itex],...v[itex]_{n}[/itex]) = w[itex]\leftrightarrow[/itex] R]

and the following restrictions are satisfied:

(i) the variable w is not free in Q

(ii) the variables v[itex]_{1}[/itex],...v[itex]_{n}[/itex],w are distinct

(iii) R has no free variables other than v[itex]_{1}[/itex],...v[itex]_{n}[/itex], w

(iv) Q and R are formulas in which the only non-logical constants are the primitive symbols and previously defined symbols of set theory

(v) the formula Q → ([itex]\exists[/itex]!)R is derivable from the axioms and preceding definitions

Such conditional definition is required to define operators or symbols which are not defined in some of its domain, e.g. division by zero a/b=c which requires c≠0.

**Question**

After giving this requirements, he continued to give example of conditionally defining the division operator by such:

(2) x/y = z[itex]\leftrightarrow[/itex](y≠0 →x=y.z)&(y=0→z=0)

and say that any conditional definition satisfying the rule stated above may be converted into a proper definition by writing it as:

(3) O( v[itex]_{1}[/itex],...v[itex]_{n}[/itex])=w[itex]\leftrightarrow[/itex](Q→R)&(-Q→w=0)

My first question is: how could we justify the change from the format shown in (1) which is an conditional implication into (3). No explanation is present in his book (including his other book on "introduction of logic"). The two formats seem very different for me.

My second question is: why do we need the (i) requirement that the variable w is not free in Q. In the next section on intersection (p.25), he gives a formal definition of intersection [itex]\cap[/itex] as:

(4) A[itex]\cap[/itex]B=C [itex]\leftrightarrow[/itex]([itex]\forall[/itex]x)(x[itex]\in[/itex]C[itex]\leftrightarrow[/itex]x[itex]\in[/itex]A&x[itex]\in[/itex]B) & C is a set

in which he says that the natural tendency to write in place of definition 5 the formula as

(5) A[itex]\cap[/itex]B=C[itex]\leftrightarrow[/itex]([itex]\forall[/itex]x)(x[itex]\in[/itex]C[itex]\leftrightarrow[/itex]x[itex]\in[/itex]A&x[itex]\in[/itex]B) is wrong because it does not translate back into general variables in a satisfactory manner, for it becomes:

(6) A,B,C are sets→ A[itex]\cap[/itex]B=C[itex]\leftrightarrow[/itex]([itex]\forall[/itex]x)(x[itex]\in[/itex]C[itex]\leftrightarrow[/itex]x[itex]\in[/itex]A&x[itex]\in[/itex]B)

which has free occurrence of variable C in its hypothesis (violating the (i) requirement).

He further asserts that the reason for preventing this free occurrence of C in the hypothesis (6) is obvious, if it is there we cannot prove for instance that 0[itex]\cap[/itex]0≠C.

Now i am completely lost here :( I understand that the translation from equation (5) to (6) is due to the implicit assumption that A,B,C are sets. But I cannot see the difference between (4) and (6). Why is it that with the free occurrence of C in 6 we cannot prove 0[itex]\cap[/itex]0≠C.

This leads to my last question. Why does 0[itex]\cap[/itex]0≠C? Isnt it true that 0[itex]\cap[/itex]0=0? And finally why do we need the additional sentence of "Z is a set" in intersection definition (4)?

**Conclusion**

There you go. My four questions are:

1. How do we justify the change of definition format from (1) to (3)

2. Why do we need the requirement of (i) the variable w is not free in Q

3. Why is it that in equation (6) which has free occurrence of C in its hypothesis we cannot prove that 0[itex]\cap[/itex]0≠C.

4. Why do we need an additional sentence of C is a set in the intersection formal definition (4):

A[itex]\cap[/itex]B=C [itex]\leftrightarrow[/itex]([itex]\forall[/itex]x)(x[itex]\in[/itex]C[itex]\leftrightarrow[/itex]x[itex]\in[/itex]A&x[itex]\in[/itex]B) & C is a set

Help would be much appreciated :D

Cheers