Theorem 1.7: (a, b ) = (c, d ) if and only if a = c and b = d.

Proof: If a = c and b = d then

{a } = {c } and {a, b } = {c, d }

so

{{a }, {a, b }} = {{c }, {c, d }}.

For the converse, assume

(a, b) = (c, d )

then

{{a }, {a, b }} = {{c }, {c, d }}

by Definition 1.8

case 1: a is not equal to b.

so

The elements of {{a }, {a , b }} are {a } and {a, b }. so either

{c } = {a }

or

{c} = { a , b}.

If {c } = {a } then since

then

in which case

c = a.

The only thing that could go wrong, then would be if the other eventuality were the case i.e. that

{c} = {a , b}

But in that case,

so

in which case

a = c

but the same argument would show that

b = c

This would give us

a = c = b

and

a = b

which is not the case here. Conclude that

{c} = {a}

and

a = c.

Next,

so

Thus either

{a , b} = {c}

or

{a , b} = {c , d}.

We can't have {a , b } = {c } because we know that a = c, but b is not equal to a so b cannot be equal to c. Thus

and so

The only other possibility would be that

{a , b } = { c , d }.

Thus

we know that

so

b = d.

case 2: a = b

In this case

{a , b } = {a , a } = {a }

So

{{a }, {a , b }} = {{a }, {a , a }} = {{a }, {a }} = {{a }}

So if

(a , b ) = (c , d )

then

{{a }} = {{c }, {c , d }}.

so

{c} = {a}

and hence

c = a.

so

{c , d} = {a}

so

so

d = a.

We then have that

c = d = a = b

so

a = c

and

b = d.

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