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Proof step 2 from proof inv-diff-n (previous step)

inv(st \ ns), n ∈ st ⊦ inv(st[n]), n ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ∈ st, ∀ n0. n0 ∈ st \ ns → # (st \ ns)[n0].list = # (st \ ns)[n0].ladrlist ∧ inv-flashsize((st \ ns)[n0].list, (st \ ns)[n0].ladrlist) ∧ ¬ dups((st \ ns)[n0].ladrlist) ⊦ n ∈ ns, inv-flashsize(st[n].list, st[n].ladrlist) ∧ ¬ dups(st[n].ladrlist)

Rule argument

Additional information

Used simplifier rules: inv-store: ⊦ inv(st) ↔ (∀ n. n ∈ st → inv(st[n]));%inv-block: ⊦ inv(ablock) ↔ # ablock.list = # ablock.ladrlist ∧ inv-flashsize(ablock.list, ablock.ladrlist) ∧ ¬ dups(ablock.ladrlist);%inv-flashsize-size: ⊦ inv-flashsize(x, lax) → (# x = # lax ↔ true);