$v$	`::=`	$() \;\;\|\;\; n \in \mathbb{N} \;\;\|\;\; l \in L \;\;\|\;\; \mu_{ptr} f. \lambda \overrightarrow{x}. t$
$b$	`::=`	$\overrightarrow{v} \;\;\|\;\; \emptyset $
$\odot$	`::=`	$+\;\;\|\;\;-\;\;\|\;\;\times$
$t$	`::=`	$v \;\;\|\;\; x \;\;\|\;\; t \odot x \;\;\|\;\; (t\;\overrightarrow{u})_{ptr}$
		$\text{alloc}\;t\;\;\|\;\;t[t]\;\;\|\;\;t[t]\gets t\;\|$
		$\text{let}\;x=t\;\text{in}\;t$
		$\text{if}\;t\;\text{then}\;t\;\text{else}\;t$


      let x = alloc 2 in
      x[0] ← 5
      x[1] ← 15
      ...
      let y = alloc 1 in
      y[0] ← x[0] + x[1]
      ... // x is not used again

Can we free `x`?


      let x = alloc 2 in
      x[0] ← 5
      x[1] ← 15
      ...
      let y = alloc 1 in
      y[0] ← x[0] + x[1]
      ...

$ \{x \mapsto [5,15] \}$

y[0] ← x[0] + x[1]

$ \{\lambda\_.\,y \mapsto 20 \} $


      let x = alloc 2 in
      x[0] ← 5
      x[1] ← 15
      ...
      let y = alloc 1 in
      y[0] ← x[0] + x[1]
      ...

$ \{x \mapsto [5,15]\,\ast\, x \gets \emptyset\}$

y[0] ← x[0] + x[1]

$ \{\lambda\_.\,y \mapsto 20\,\ast\,\color{red}{??}\} $

$ \{x \mapsto [5,15]\,\ast\, x \gets \emptyset\}$

y[0] ← x[0] + x[1]

$ \{\lambda\_.\,y \mapsto 20\,\ast\,\diamond\,2\,\ast\,\dagger\!\!\dagger\!\!\{x\}\} $

What about the stack?

The permission to place $x$ in the stack
This permission can be shared $\text{Stackable}_{p+q}\;x\;\equiv \text{Stackable}_p\;x\;\ast\;\text{Stackable}_q\;x$
$p$ is a fractional number $(0..1]$
In particular, $ \text{Stackable}\;1\;x$ ensures x is not in the sack