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Proof step 18 from proof inv-alloc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
st[n, mkblock(x, lax, m)][n0] = mkblock(x0, lax0, m0), inv-flashsize(x, lax), 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), n0 = n ∨ ¬ n0 ∈ ns ⊦ dups(lax), inv-flashsize(x0, lax0) ∧ ¬ dups(lax0), n0 ≠ n ∧ n0 ∈ ns

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ ns -- n0 ↔ n ≠ n0 ∧ n ∈ ns; (s)% ⊦ ¬ n ∈ ns → (st \ ns)[n] = st[n]; (s)