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Proof step 11 from proof inv-alloc (previous step)

∀ 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 ∈ st[n, mkblock(x0, lax0, m)] \ (ns -- n), inv-flashsize(x0, lax0) ∧ ¬ dups(lax0)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Proof step 12:
n ∈ st[n, mkblock(x0, lax0, 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 ∈ st[n, mkblock(x0, lax0, m)] \ (ns -- n), inv-flashsize(x0, lax0) ∧ ¬ dups(lax0)

Rule argument

fma: n ∈ st[n, mkblock(x0, lax0, m)]

Additional information

Used simplifier rules: ⊦ n ∈ st[n0, ablock] ↔ n = n0 ∨ n ∈ st; (ws) Used formulas: n ∈ st[n, mkblock(x0, lax0, m)]