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

Proof: Suppose towards contradiction that X $\subset$ $\mathbb{N}$ has 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 $\mathbb{N}$, 1 is not 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.

12. Statement: If *y* is not free in ϕ, using the ∃-introduction rule returns an invalid deduction.

Proof Attempts:

1. ϕ=∀y(xЄy→yЄx)

ϕ[y|x]=∀x(xЄy→yЄx) true

∀y∃x(xЄy→yЄx) true

2. ϕ=∀y(x = ¬x)

ϕ[y|x]=∀x(x = ¬x) false

3. ϕ=∀y(y = ¬x)

ϕ[y|x]=∀x(y = ¬x) true

∀y∃x(y = ¬x) true

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 $\mathbb{N}$

14. Statement: If * X* is a finite set and

*is an injective function from*

**f***→*

**X***, then*

**X***is bijective.*

**f**Proof: Suppose towards contradiction that

*is injective but not surjective and thus not bijective. There then must be a point yЄ*

**f***such that there is no point zЄ*

**X***with f(z)=y. Since*

**X***is a function, every xЄ*

**f***must be the x-coordinate in the relation*

**X***. Hence, we must have some z*

**f**_{1}≠z

_{2}with

*(x*

**f**_{1}) =

*(z*

**f**_{2}) which is not possible. Thus,

*must be bijective.*

**f**