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I'm creating this thread for any major difficulties I come across in How to Prove It: A Structured Approach, 2nd edition by Daniel J. Velleman. This is a self-study (with any assistance if I can get it).
The problem is to analyze the logical form of the given statement, using only the symbols [itex]\in[/itex], [itex]\notin[/itex], [itex]=[/itex], [itex]≠[/itex], [itex]\wedge[/itex], [itex]\vee[/itex], [itex]\rightarrow[/itex], [itex]\leftrightarrow[/itex], [itex]\forall[/itex], and [itex]\exists[/itex] in our answer, but not [itex]\subseteq[/itex], is-not-[itex]\subseteq[/itex], [itex]\wp[/itex], [itex]\cup[/itex], [itex]\cap[/itex], \, [itex]\{[/itex], [itex]\}[/itex], or [itex]\neg[/itex]. (Thus, as the text goes on to say, we must write out the definitions of some set theory notation, and we must use equivalences to get rid of any occurrences of [itex]\neg[/itex].) This is Exercise 1-(c) on page 81.
The statement is [itex]\{n^2 + n + 1\ |\ n \in \mathbb N\} \subseteq \{2n + 1\ |\ n \in \mathbb N\}[/itex].
Although it seems correct to me, it's not exactly the solution given in the Appendix of solutions. His answer seems to be derived from facts that I can't find or discern from any of the previous discussion. What I ideally want is to see his solution derived in a way that I should've known from what's been explained in the text so far. If not that then one or two standard definitions or logical equivalences that might or might not have been introduced but nonetheless work to get me to the same result.
My answer:
His answer:
Homework Statement
The problem is to analyze the logical form of the given statement, using only the symbols [itex]\in[/itex], [itex]\notin[/itex], [itex]=[/itex], [itex]≠[/itex], [itex]\wedge[/itex], [itex]\vee[/itex], [itex]\rightarrow[/itex], [itex]\leftrightarrow[/itex], [itex]\forall[/itex], and [itex]\exists[/itex] in our answer, but not [itex]\subseteq[/itex], is-not-[itex]\subseteq[/itex], [itex]\wp[/itex], [itex]\cup[/itex], [itex]\cap[/itex], \, [itex]\{[/itex], [itex]\}[/itex], or [itex]\neg[/itex]. (Thus, as the text goes on to say, we must write out the definitions of some set theory notation, and we must use equivalences to get rid of any occurrences of [itex]\neg[/itex].) This is Exercise 1-(c) on page 81.
Homework Equations
The statement is [itex]\{n^2 + n + 1\ |\ n \in \mathbb N\} \subseteq \{2n + 1\ |\ n \in \mathbb N\}[/itex].
The Attempt at a Solution
Although it seems correct to me, it's not exactly the solution given in the Appendix of solutions. His answer seems to be derived from facts that I can't find or discern from any of the previous discussion. What I ideally want is to see his solution derived in a way that I should've known from what's been explained in the text so far. If not that then one or two standard definitions or logical equivalences that might or might not have been introduced but nonetheless work to get me to the same result.
My answer:
[itex]\{n^2 + n + 1\ |\ n \in \mathbb N\} \subseteq \{2n + 1\ |\ n \in \mathbb N\}[/itex]
[itex]\equiv \forall x(x \in \{n^2 + n + 1\ |\ n \in \mathbb N\} \rightarrow x \in \{2n + 1\ |\ n \in \mathbb N\})[/itex]
[itex]\equiv \forall x(x \in \{y\ |\ \exists n \in \mathbb N (y = n^2 + n + 1)\} \rightarrow x \in \{y\ |\ \exists n \in \mathbb N (y = 2n + 1)\})[/itex]
[itex]\equiv \forall x(\exists n \in \mathbb N (x = n^2 + n + 1) \rightarrow \exists n \in \mathbb N (x = 2n + 1)).[/itex]
His answer:
[itex]. . . \equiv \forall n \in \mathbb N \exists m \in \mathbb N (n^2 + n + 1 = 2m + 1).[/itex]
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