The uploaded image contains a proof of a mathematical statement regarding vector spaces. I'll analyze and fill in the blanks in the proof.

Proof Analysis:

The statement to prove is:
Let $V$ be any vector space and let $\mathbf{x}, \mathbf{y}, \mathbf{z} \in V$. Prove that if $\mathbf{x} + \mathbf{z} = \mathbf{y} + \mathbf{z}$, then $\mathbf{x} = \mathbf{y}$.

Missing Steps:

1. First Blank:
   $\mathbf{z} + (-\mathbf{z}) = 0$, because additive inverses exist in $V$ (a property of vector spaces).

2. Second Blank:
   $\mathbf{x} + (\mathbf{z} + (-\mathbf{z})) = \mathbf{y} + (\mathbf{z} + (-\mathbf{z}))$, because addition in $V$ is associative.

3. Third Blank:
   $\mathbf{x} + 0 = \mathbf{y} + 0$, because $\mathbf{z} + (-\mathbf{z}) = 0$, the additive identity.

Completed Proof:

1. Given $\mathbf{x} + \mathbf{z} = \mathbf{y} + \mathbf{z}$, suppose $V$ is a vector space.
2. There exists a vector $-\mathbf{z} \in V$ such that $\mathbf{z} + (-\mathbf{z}) = 0$ (additive inverse).
3. Adding $-\mathbf{z}$ to both sides of $\mathbf{x} + \mathbf{z} = \mathbf{y} + \mathbf{z}$, we have: $(\mathbf{x} + \mathbf{z}) + (-\mathbf{z}) = (\mathbf{y} + \mathbf{z}) + (-\mathbf{z})$
4. Using associativity of addition: $\mathbf{x} + (\mathbf{z} + (-\mathbf{z})) = \mathbf{y} + (\mathbf{z} + (-\mathbf{z}))$
5. Since $\mathbf{z} + (-\mathbf{z}) = 0$, we get: $\mathbf{x} + 0 = \mathbf{y} + 0$
6. Adding the additive identity (0) does not change a vector, so: $\mathbf{x} = \mathbf{y}.$

Thus, $\mathbf{x} + \mathbf{z} = \mathbf{y} + \mathbf{z}$ implies $\mathbf{x} = \mathbf{y}$, and $V$ has the cancellation property.

Do you have any further questions or need clarification on any part of this proof?

Here are 5 related questions for further exploration:

1. What are the defining properties of a vector space?
2. Why is the existence of additive inverses critical for vector space operations?
3. How does the cancellation property relate to linear independence in vector spaces?
4. Can this proof be extended to infinite-dimensional vector spaces? Why or why not?
5. How would this proof change in a non-commutative algebraic structure?

Tip: Always verify the properties of a mathematical structure (e.g., vector space axioms) before applying results like the cancellation property.