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Proof step 9 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 ∈ ns, inv-flashsize(x0, lax0) ∧ ¬ dups(lax0)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Proof step 10:
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)

Rule argument

fma: n0 = n

Additional information

Used simplifier rules: ⊦ st[n, ablock][n] = ablock; (s)%⊦ mkblock(x, lax, m) = mkblock(x0, lax0, m0) ↔ x = x0 ∧ lax = lax0 ∧ m = m0; (ws)% ⊦ n ∈ st[n, ablock]; (s) Used formulas: n0 = n