- #1

bobby2k

- 127

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I am supposed to prove:

If A [itex]\neq[/itex] [itex]\phi[/itex] and B [itex]\neq[/itex] [itex]\phi[/itex] then

A[itex]\times[/itex] B [itex]\neq[/itex] [itex]\phi[/itex]

The HINT in the back of the book gives:

A [itex]\neq[/itex] [itex]\phi[/itex] [itex]\wedge[/itex] B [itex]\neq[/itex] [itex]\phi[/itex]

[itex]\Rightarrow[/itex] [itex]\exists[/itex]a[itex]\subseteq[/itex]A [itex]\wedge[/itex]b[itex]\subseteq[/itex] B so that (a,b) [itex]\subseteq[/itex] A[itex]\times[/itex] B

I have 2 questions

1.Is it enough for the proof to write the hint, and then say "hence A [itex]\times[/itex]B" is not empty.

2. Or do I have to prove the hint?

If I have to prove the hint, is this a correct way to prove it?

[itex]A \neq \phi \wedge B \neq \phi [/itex] -Premise-(1)

[itex]A \neq \phi [/itex] - 1 and simplification-(2)

[itex]B \neq \phi [/itex] - 1 and simplification-(3)

[itex] \exists a \in A[/itex] - 2 and since A is not empty it has atleast one element-(4)

[itex] \exists b \in B[/itex] - 3 and since B is not empty it has atleast one element-(5)

[itex] (\exists a \in A) \wedge (\exists b \in B)[/itex]- 4 and 5- (6)

[itex](a,b) \in A \times B[/itex] - 6, and definition of [itex]A \times B [/itex] -(7)

[itex] A \times B \neq \phi [/itex] - 7 and definition of [itex]\phi[/itex] (8)

Is step (4) and (5) ok? I haven't read an axiom that specifies that every non-empty set has ateleast one element. Do I maybe have to use the face that there is only one empty set. And since there is only one empty set, every other set must have atleast one element, or else they would be the empty set?

Should step (7) have more explanation?

Also should existencial generalistion or specification have been cited anywhere in the proof?

If A [itex]\neq[/itex] [itex]\phi[/itex] and B [itex]\neq[/itex] [itex]\phi[/itex] then

A[itex]\times[/itex] B [itex]\neq[/itex] [itex]\phi[/itex]

The HINT in the back of the book gives:

A [itex]\neq[/itex] [itex]\phi[/itex] [itex]\wedge[/itex] B [itex]\neq[/itex] [itex]\phi[/itex]

[itex]\Rightarrow[/itex] [itex]\exists[/itex]a[itex]\subseteq[/itex]A [itex]\wedge[/itex]b[itex]\subseteq[/itex] B so that (a,b) [itex]\subseteq[/itex] A[itex]\times[/itex] B

I have 2 questions

1.Is it enough for the proof to write the hint, and then say "hence A [itex]\times[/itex]B" is not empty.

2. Or do I have to prove the hint?

If I have to prove the hint, is this a correct way to prove it?

[itex]A \neq \phi \wedge B \neq \phi [/itex] -Premise-(1)

[itex]A \neq \phi [/itex] - 1 and simplification-(2)

[itex]B \neq \phi [/itex] - 1 and simplification-(3)

[itex] \exists a \in A[/itex] - 2 and since A is not empty it has atleast one element-(4)

[itex] \exists b \in B[/itex] - 3 and since B is not empty it has atleast one element-(5)

[itex] (\exists a \in A) \wedge (\exists b \in B)[/itex]- 4 and 5- (6)

[itex](a,b) \in A \times B[/itex] - 6, and definition of [itex]A \times B [/itex] -(7)

[itex] A \times B \neq \phi [/itex] - 7 and definition of [itex]\phi[/itex] (8)

Is step (4) and (5) ok? I haven't read an axiom that specifies that every non-empty set has ateleast one element. Do I maybe have to use the face that there is only one empty set. And since there is only one empty set, every other set must have atleast one element, or else they would be the empty set?

Should step (7) have more explanation?

Also should existencial generalistion or specification have been cited anywhere in the proof?

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