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A well-pruned set-theoretic tree of height $\kappa$ has the property that every node $t$ extends to every higher level below $\kappa$: if $\mathrm{ht}(t)<\beta <\kappa$, there is $u$ of height $\beta$ with $t{<}_{T}u$. Equivalently, the heights of extensions of every node are unbounded in $\kappa$, since taking predecessors then gives an extension at any prescribed intermediate level. This is stronger than merely having no terminal nodes. For a kappa-tree we use the usual regular uncountable height cardinal and levels of size less than that cardinal.

Prune an ${\aleph }_1$-Suslin tree as in part (a), then use its nodes as forcing conditions, with extensions stronger. Two conditions are compatible exactly when comparable, so the absence of uncountable tree antichains is the forcing countable chain condition for forcing. For each $\alpha <{\omega }_1$, the set $D_{\alpha }$ of nodes of height at least $\alpha$ is dense, by well-pruned set-theoretic tree.

If $2^{{\aleph }_0}>{\aleph }_1$, full Martin's axiom includes ${\mathrm{MA}}_{{\aleph }_1}$. It would provide a filter in an ordered set meeting all these dense subsets of a forcing order. Directedness makes that filter in an ordered set a chain in a partial order, and meeting every $D_{\alpha }$ makes its heights unbounded, producing an uncountable branch. This contradicts the Suslin-tree property. A Suslin set-theoretic tree together with failure of Continuum hypothesis therefore implies failure of Martin's axiom. This is the Suslin-tree obstruction to Martin's axiom.

A well-pruned set-theoretic tree that is an Aronszajn tree and a Suslin tree gives a forcing with the countable chain condition for forcing: stronger nodes extend weaker ones. The ${\omega }_1$ dense subsets of a forcing order of nodes at or above each level cannot all be met by a filter in an ordered set, since that would produce a cofinal branch. Thus ${\mathrm{MA}}_{{\aleph }_1}$ fails. When the continuum exceeds ${\aleph }_1$, full Martin axiom includes this instance.