13. Statement: 5^{2n+1}+2^{2n+1} is divisible by 7 for all n Є $\mathbb{N}$.

Proof: We can begin by proving that 5^{2n+1}+2^{2n+1} being divisible by 7 is true for the base case, n = 0.

5^{2(0)+1}+2^{2(0)+1} =

5^{1}+2^{1}=

5+2=7

7/7 =1

Based on this idea, assume the argument holds true for n=k.

5^{2k+1}+2^{2k+1}=7J where JЄ$\mathbb{N}$

2^{2k+1}=7J-5^{2k+1}

Next, we must show that the argument holds for n=k+1.

5^{2(k+1)+1}+2^{2(k+1)+1}=5^{2k+3}+2^{2k+3}

=5^{2k+1}(5^{2})+2^{2k+1}(2^{2})

=25(5^{2k+1})+4(2^{2k+1})

=25(5^{2k+1})+4(7J-5^{2k+1})

=25(5^{2k+1})+28J-4(5^{2k+1})

=21(5^{2k+1})+28J

=7[3(5^{2k+1})+4J]

7[3(5^{2k+1})+4J] will always be divisible by 7 regardless of the value of k as long as k Є $\mathbb{N}$. Hence, 5^{2n+1}+2^{2n+1} is divisible by 7 for all n Є N. ∎

12. Statement: If y is not free in ϕ=∀y(x=y), using the ∃-introduction rule returns an invalid deduction.

Proof: 1. ⊢∀y(x=y)

2. ⊢∀y(y=y) from 1 by ϕ[y|x]

3. ⊢∃x∀y(x=y) from 1 by ∃I

However, ⊢∃x∀y(x=y) is not a valid deduction because there does not exist an x that for all y x=y. ∎

11. Statement: $\mathbb{N}$ is well-ordered.

Proof: Let X ⊂ $\mathbb{N}$ have no least element. If 0 Є $\mathbb{N}$, then X will have a least element and we would be done. Now suppose 0 is not an element in $\mathbb{N}$, thus, 1 cannot be an element in $\mathbb{N}$, and n-1 is not an element in $\mathbb{N}$. If n Є X under these assumptions, then X would have a least element. So, n is not an element in $\mathbb{N}$ for every n Є $\mathbb{N}$. Therefore, X=∅, but $\mathbb{N}$ is not the empty set. Hence, $\mathbb{N}$ has a least element and is thus well-ordered. ∎

(Prof Simmons, 2-25-15): This is on the right track (showing that a subset of $\mathbb{N}$ with no least element is empty). The induction is a bit unclear: 1 and $n-1$ not belonging to $X$ isn't sufficient to show $n\notin X$; you need that every element less than $n$ not belong to $X$. You should also explain what $n$ is (an arbitrary natural number used for induction) when you first use it. The last line is problematic; having a least element doesn't make a set well ordered.

Comments - Andrew Furash

For 12, you have the right idea, but your number 1 is already an incorrect statement. You need to start with something that is correct and conclude with something incorrect.

13 looks good!