TITLE: Raising the index of accessibility
QUESTION [17 upvotes]: In the standard reference books Locally presentable and accessible categories  (Adamek-Rosicky, Theorem 2.11) and Accessible categories (Makkai-Pare, $\S$2.3), it is shown that for regular cardinals $\lambda\le\mu$, the following are equivalent:

Every $\lambda$-accessible category is $\mu$-accessible.
For every $\mu'<\mu$, the set $P_\lambda(\mu')$ of subsets of $\mu'$ of cardinality $<\lambda$ has a cofinal subset of cardinality $<\mu$.

This relation is denoted $\lambda\unlhd \mu$ (or $\lambda \lhd \mu$ for the irreflexive version).  
In Higher Topos Theory (Lurie, Definition A.2.6.3) the relation $\lambda\ll\mu$ is defined to mean

For every $\lambda'<\lambda$ and $\mu'<\mu$ we have $(\mu')^{\lambda'} <\mu$.

By Example 2.13(4) of Adamek-Rosicky, if $\lambda\ll\mu$ and $\lambda<\mu$ then $\lambda\lhd \mu$.  (Indeed, in this case $P_\lambda(\mu')$ itself has cardinality $<\mu$.)
Does the converse hold?  That is, if $\lambda\lhd \mu$ do we have $\lambda\ll\mu$?
Note that the two relations definitely differ in the reflexive case: $\lambda\unlhd\lambda$ is always true, but $\lambda\ll\lambda$ holds only when $\lambda$ is inaccessible.  It also seems that the converse implies the generalized continuum hypothesis for regular cardinals, since if $\lambda^+ < 2^\lambda$ then we have $\lambda^+ \lhd \lambda^{++}$ but not $\lambda^+ \ll \lambda^{++}$.  Thus, the converse is not provable in ZFC.  Is it disprovable?

REPLY [7 votes]: More generally, the following is Fact 2.5 of Lieberman, Rosický, and Vasey - Internal sizes in $\mu$-abstract elementary classes:
Theorem. Assume $\lambda$ and $\mu$ are regular cardinals and $2^{<\lambda} < \mu$. Then $\lambda \triangleleft \mu$ if and only if $\lambda \ll \mu$.
Assuming GCH, $2^{<\lambda} = \lambda$, so we recover Gabe's answer (the proof is the same).