We prove the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$ and $(1) \Rightarrow (4) \Rightarrow (5) \Rightarrow (6) \Rightarrow (1)$.

(1) $\Rightarrow$ (2): We prove by induction on $n$ that a connected acyclic graph with $n$ vertices has $n-1$ edges.

Base case: $n=1$ gives 0 edges. ✓

Inductive step: Let $T$ be connected and acyclic with $n \geq 2$ vertices. Since $T$ is connected, every vertex has degree at least 1. By the handshaking lemma, $\sum \deg(v) = 2m$ where $m$ is the number of edges. If all degrees were at least 2, we'd have $2m \geq 2n$, so $m \geq n$.

But a connected graph with $n$ vertices and at least $n$ edges must contain a cycle (by pigeonhole on any spanning tree). This contradicts acyclicity. Therefore, some vertex $v$ has degree 1.

Removing $v$ and its incident edge yields $T' = T - v$, which is connected and acyclic with $n-1$ vertices. By induction, $T'$ has $(n-1)-1 = n-2$ edges. Thus $T$ has $(n-2) + 1 = n-1$ edges.

(2) $\Rightarrow$ (3): Suppose $T$ is connected with $n-1$ edges. If $T$ contained a cycle $C$, removing any edge from $C$ would leave $T$ connected (can route around the cycle). This contradicts minimality: a connected graph requires at least $n-1$ edges. Thus $T$ is acyclic.

(3) $\Rightarrow$ (1): Let $T$ be acyclic with $n$ vertices and $n-1$ edges. Suppose $T$ has $k$ connected components, each a tree. If component $i$ has $n_i$ vertices, it has $n_i - 1$ edges (by (1)$\Rightarrow$(2) applied to each component). Summing: $n-1 = \sum_{i=1}^k (n_i - 1) = n - k$ Thus $k = 1$, so $T$ is connected.

(1) $\Rightarrow$ (4): Suppose $T$ is connected and acyclic. Removing any edge $e$ disconnects $T$: otherwise, there would be a path avoiding $e$ between $e$'s endpoints, creating a cycle. Thus $T$ is minimally connected.

(4) $\Rightarrow$ (5): Let $T$ be minimally connected. Then $T$ is acyclic (else removing an edge from a cycle maintains connectivity). Adding any edge $\{u,v\}$ creates exactly one cycle: the existing path from $u$ to $v$ plus the new edge. Thus $T$ is maximally acyclic.

(5) $\Rightarrow$ (6): If some pair $u,v$ had no path, adding edge $\{u,v\}$ wouldn't create a cycle, contradicting maximality. If some pair had two distinct paths, these paths form a cycle, contradicting acyclicity. Thus every pair has exactly one path.

(6) $\Rightarrow$ (1): Unique paths imply connectivity. If a cycle existed, its vertices would have two distinct paths (clockwise and counterclockwise around the cycle), contradicting uniqueness. Thus $T$ is connected and acyclic.