[proofplan] Assume toward a contradiction that $T$ is a normal Suslin tree. We use $T$ itself as a forcing poset, ordered so that stronger conditions are higher extensions in the tree. The Suslin antichain condition makes this forcing ccc, and normality gives one dense set $D_\alpha$ requiring the generic filter to reach height at least $\alpha$ for each $\alpha < \omega_1$. Applying $MA_{\aleph_1}$ produces a filter meeting all these dense sets; directedness then forces the filter to be a chain, and meeting all heights makes that chain uncountable, contradicting that $T$ is Suslin. [/proofplan] [step:Turn the Suslin tree into a ccc forcing poset] Suppose, toward a contradiction, that $T$ is a normal Suslin tree. For each node $t \in T$, let $\operatorname{ht}_T(t) < \omega_1$ denote the height of $t$ in the tree. We regard $T$ as a forcing poset $\mathbb{P} = (T, \le_{\mathbb{P}})$ by declaring that $s \le_{\mathbb{P}} t$ if and only if $s$ extends $t$ in the tree order. Thus stronger forcing conditions are higher nodes of the tree. An antichain in $\mathbb{P}$ is exactly a set of pairwise incompatible nodes of $T$. Indeed, if two nodes $p,q \in T$ have a common stronger condition $r \in T$, then $r$ extends both $p$ and $q$, so $p$ and $q$ both lie in the predecessor set of $r$; that predecessor set is linearly ordered by the definition of a tree, hence $p$ and $q$ are comparable. Conversely, comparable nodes have the higher node as a common stronger condition. Since $T$ is Suslin, it has no uncountable antichain. Hence every antichain in $\mathbb{P}$ is countable, so $\mathbb{P}$ is ccc. [/step] [step:Define dense sets forcing the filter to reach every countable height] For each ordinal $\alpha < \omega_1$, define $D_\alpha := \{t \in T : \operatorname{ht}_T(t) \geq \alpha\}$. We claim that $D_\alpha$ is dense in $\mathbb{P}$. Let $t \in T$ be arbitrary. Since $T$ is normal and has height $\omega_1$, there is an extension $s \in T$ of $t$ with $\operatorname{ht}_T(s) \geq \alpha$. By the definition of $\le_{\mathbb{P}}$, this means $s \le_{\mathbb{P}} t$, and by construction $s \in D_\alpha$. Therefore every condition in $\mathbb{P}$ has a stronger extension in $D_\alpha$, so $D_\alpha$ is dense. [guided] We are working with the forcing poset $\mathbb{P} = (T, \le_{\mathbb{P}})$ whose conditions are nodes of $T$, with $s \le_{\mathbb{P}} t$ exactly when $s$ extends $t$ in the tree order. For each $\alpha < \omega_1$, we want a dense set whose only purpose is to force the eventual filter to contain some node at height at least $\alpha$. Define $D_\alpha := \{t \in T : \operatorname{ht}_T(t) \geq \alpha\}$. We verify density carefully. A subset $D \subseteq \mathbb{P}$ is dense if for every condition $t \in \mathbb{P}$ there exists $s \in D$ such that $s \le_{\mathbb{P}} t$. Here $s \le_{\mathbb{P}} t$ means that $s$ is a tree-extension of $t$, so we must find an extension of $t$ whose height is at least $\alpha$. This is exactly where normality of the tree is used. Since $T$ is normal and has height $\omega_1$, every node of $T$ has extensions arbitrarily high below $\omega_1$. In particular, given $t \in T$ and $\alpha < \omega_1$, there exists a node $s \in T$ extending $t$ with $\operatorname{ht}_T(s) \geq \alpha$. Therefore $s \in D_\alpha$ and $s \le_{\mathbb{P}} t$. Hence $D_\alpha$ is dense in $\mathbb{P}$. [/guided] [/step] [step:Apply $MA_{\aleph_1}$ to obtain a filter meeting all height requirements] The family $\{D_\alpha : \alpha < \omega_1\}$ has cardinality $\aleph_1$, and $\mathbb{P}$ is ccc. By Martin's Axiom at $\aleph_1$, there exists a forcing filter $G \subseteq \mathbb{P}$ such that $G \cap D_\alpha \neq \varnothing$ for every $\alpha < \omega_1$. Thus for every $\alpha < \omega_1$, choose a node $t_\alpha \in G \cap D_\alpha$; by the definition of $D_\alpha$, this node satisfies $\operatorname{ht}_T(t_\alpha) \geq \alpha$. [/step] [step:Use directedness of the filter to show it is a chain] Let $p,q \in G$. Since $G$ is a filter on $\mathbb{P}$, it is directed: there exists $r \in G$ such that $r \le_{\mathbb{P}} p$ and $r \le_{\mathbb{P}} q$. By the definition of $\le_{\mathbb{P}}$, the node $r$ extends both $p$ and $q$ in the tree. Therefore $p$ and $q$ both lie on the unique predecessor chain below $r$, and predecessor chains in a tree are linearly ordered. Hence $p$ and $q$ are comparable in the tree order. Since $p,q \in G$ were arbitrary, $G$ is a chain in $T$. [/step] [step:Derive an uncountable branch and contradict the Suslin property] For every $\alpha < \omega_1$, choose $t_\alpha \in G \cap D_\alpha$. Then $\operatorname{ht}_T(t_\alpha) \geq \alpha$. Thus the set of heights $H := \{\operatorname{ht}_T(t) : t \in G\}$ is unbounded in $\omega_1$. Every bounded subset of $\omega_1$ is countable, so $H$ is uncountable. Since a chain in a tree contains at most one node on each level, the chain $G$ itself is uncountable. We have constructed an uncountable chain in $T$, contradicting the assumption that $T$ is a Suslin tree. Therefore no normal Suslin tree exists under $MA_{\aleph_1}$. [/step]