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Proof step 5 from proof size-in (previous step)

a ∈ st, st = (st -- a)[a, st[a]] ⊦ # st[a, d] = # st

User applied rule insert equation

And obtained the following new goals

Proof step 6:
a ∈ (st -- a)[a, st[a]] ⊦ # (st -- a)[a, st[a]][a, d] = # (st -- a)[a, st[a]]

Rule argument

pos: left 2 flags: discard

Additional information