Easton’s Lemma in forcing Easton essentially proved the following: 

Suppose $\mathbb P$ is $\kappa$-c.c., $\mathbb Q$ is $\kappa$-distributive, and $\Vdash_{\mathbb Q} \mathbb P$ is $\kappa$-c.c.  Then $\Vdash_{\mathbb P} \mathbb Q$ is $\kappa$-distributive.

Is the converse true?  If $\mathbb P$ preserves the $\kappa$-distributivity of $\mathbb Q$, does $\mathbb Q$ preserve the $\kappa$-c.c. of $\mathbb P$?
 A: Here is a counterexample using a Mahlo cardinal.  Let $\kappa$ be Mahlo, and let $\mathbb P$ be the Easton-support product of $\mathrm{Add}(\alpha)$ for $\alpha<\kappa$ inaccessible.  $\mathbb P$ is $\kappa$-c.c.  Let $\mathbb Q$ be the partial order for killing the Mahloness of $\kappa$ by shooting a club through the singular cardinals below $\kappa$, where the conditions are closed bounded subsets of $\kappa$ consisting of singular cardinals.  $\mathbb Q$ has the property that if $\delta<\kappa$, and $q \in \mathbb Q$ is a condition with supremum greater than $\delta$, then $\mathbb Q \restriction q$ is $\delta^+$-closed.
$\mathbb Q$ forces that $\mathbb P$ is no longer $\kappa$-c.c., because if $C \subseteq \kappa$ is the generic club of singulars, then $C$ is an Easton subset of $\kappa$, and all initial segments are in the ground model.  We can use $C$ to find a length-$\kappa$ chain in $\mathbb P$ and branch off from it to find a size-$\kappa$ antichain.
Now we argue that $\mathbb P$ preserves the distributivity of $\mathbb Q$.  Let $\delta < \kappa$, $\theta$ an ordinal, and suppose $f : \delta \to \theta$ is a function in $V[G][H]$, where $G \times H$ is $\mathbb P \times \mathbb Q$-generic.  Let $(p,q) \in G \times H$ be such that $q$ has supremum $>\delta$.  $\mathbb P$ factors into a product $\mathbb P_0 \times \mathbb P_1$, where $\mathbb P_1$ is $\delta^+$-closed and $| \mathbb P_0 | \leq \delta$.  Let $G_0 \times G_1$ be the factoring of $G$.  Since $\mathbb P_1 \times \mathbb Q \restriction q$ is $\delta^+$-closed, it is $\delta^+$-distributive in $V[G_0]$.  Thus $f \in V[G_0] \subseteq V[G]$.
