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Proof step 9 from proof nodes-alloc (previous step)

⊦ n ∈ st, (m0 ∈ st[n, mkblock([], [], m)] ∧ @(m1, m2) ∈ st[n, mkblock([], [], m)][m0].ladrlist → @(m0, m1, m2) ∈ st) ∧ ( m0 ∈ st[n, mkblock([], [], m)] ∧ @(m1, m2) ∈ st[n, mkblock([], [], m)][m0].ladrlist → st[n, mkblock([], [], m)][@(m0, m1, m2)] = nodes(st)[@(m0, m1, m2)])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
st[n, mkblock([], [], m)][m0] = mkblock(x, lax, m3) ⊦ n ∈ st, (m0 ∈ st[n, mkblock([], [], m)] ∧ @(m1, m2) ∈ lax → @(m0, m1, m2) ∈ st) ∧ (m0 ∈ st[n, mkblock([], [], m)] ∧ @(m1, m2) ∈ lax → st[n, mkblock([], [], m)][@(m0, m1, m2)] = nodes(st)[@(m0, m1, m2)])

Rule argument

spec: ablock name: elim seq: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m) subst: [ablock, x, lax, m] → [st[n, mkblock([], [], m)][m0], x, lax, m3]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st[n, mkblock([], [], m)][m0], x, lax, m3