Does Laver Forcing add an infinitely often equal real?

Given a model $V$ of set theory and an inner model $W \subseteq V$, a real $x \in V \cap \omega^\omega$ is infinitely often equal over $W$ if for each real $y \in \omega^\omega$ there are infinitely $n$ with $x(n) = y(n)$. Cohen reals have this property and it was an open question for a while as to whether every forcing adding an infinitely often equal real also adds a Cohen real. In fact this is false.

What I want to know is whether it is ever the case that Laver forcing adds a real which is infinitely often equal over the ground model. I expect the answer to be no given the aforementioned question about finding infinitely often equal reals that are not Cohen. However, I'm struggling to prove it.

For reference, recall that Laver forcing, $\mathbb L$, is defined to be the set of trees on $\omega$ with a distinguished stem and infinitely branching at every node above the stem. A Laver real is defined in the generic extension to be the union of the stems of the trees in the generic. It's not hard to see that such a real is always dominating over the ground model. Moreover it is well known (and proved in Jech) that $\mathbb L$ never adds Cohen reals over the ground model, even when iterated.

The answer is no. This follows easily from the Laver property. Every new real which is bounded by a ground model real (say: by the identity function) is contained in a small slalom from the ground model, hence will be eventually different from lots of ground model reals.

• I figured it was something like this. Thanks. Dec 7, 2017 at 22:11
• Sorry one clarifying question: since any infinitely often equal real isn't bounded by any ground model real, how exactly does one apply the Laver property in this case? Dec 7, 2017 at 22:26
• If $x$ is infinitely equal to all reals from the ground model, then $x':=\min(x,id)$ is infinitely equal to all ground model reals bounded by $id$, and $x'$ is now bounded. Dec 7, 2017 at 22:37