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Proof step 6 from proof inv-remove-refsize0-diff (previous step)

st[n] = mkblock(x, lax, m), n ∈ st, ∀ n. n ∈ st \ ns → # (st \ ns)[n].list = # (st \ ns)[n].ladrlist ∧ inv-flashsize((st \ ns)[n].list, (st \ ns)[n].ladrlist) ∧ ¬ dups((st \ ns)[n].ladrlist), ∀ n. n ∈ dom remove-refsize0(st) ∩ ns ↔ n ∈ ∅ ⊦ m = 0, inv-flashsize(x, lax) ∧ ¬ dups(lax)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
st[n] = mkblock(x, lax, m), n ∈ st, ∀ n. n ∈ st \ ns → # (st \ ns)[n].list = # (st \ ns)[n].ladrlist ∧ inv-flashsize((st \ ns)[n].list, (st \ ns)[n].ladrlist) ∧ ¬ dups((st \ ns)[n].ladrlist), ∀ n0. ¬ n0 ∈ dom remove-refsize0(st) ∩ ns ⊦ m = 0, inv-flashsize(x, lax) ∧ ¬ dups(lax)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n ∈ ∅; (s)