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Proof step 10 from proof inv-reduce (previous step)

st[n, mkblock(x, lax, m)][n0] = mkblock(x0, lax0, m0), n0 ∈ st[n, mkblock(x, lax, m)], ∀ 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(x, lax) ∧ ¬ dups(lax) ⊦ n0 = n, n0 ∈ ns, inv-flashsize(x0, lax0) ∧ ¬ dups(lax0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
st[n0] = mkblock(x0, lax0, m0), n0 ∈ 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(x, lax) ∧ ¬ dups(lax) ⊦ n0 = n, n0 ∈ ns, inv-flashsize(x0, lax0) ∧ ¬ dups(lax0)

Rule argument

Additional information

Used simplifier rules: ⊦ n ≠ n0 → (n ∈ st[n0, ablock] ↔ n ∈ st); (s)% ⊦ n ≠ n0 → st[n0, ablock][n] = st[n]; (s)