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Proof step 11 from proof nodes-alloc-append (previous step)

⊦ adr = @(n, m, m0), n ∈ st, (adr ∈ st[n, mkblock(a ', @(m, m0) ', m1)] ↔ adr ∈ nodes(st)[@(n, m, m0), a]) ∧ (adr ∈ st[n, mkblock(a ', @(m, m0) ', m1)] → st[n, mkblock(a ', @(m, m0) ', m1)][adr] = nodes(st)[@(n, m, m0), a][adr])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ adr = @(n, m, m0), n ∈ st, (adr ∈ st[n, mkblock(a ', @(m, m0) ', m1)] → adr ∈ st) ∧ (adr ∈ st[n, mkblock(a ', @(m, m0) ', m1)] → st[n, mkblock(a ', @(m, m0) ', m1)][adr] = nodes(st)[adr])

Rule argument

Additional information

Used simplifier rules: in-nodes: ⊦ adr ∈ nodes(st) ↔ adr ∈ st;% ⊦ adr ≠ adr0 → ast[adr0, a0][adr] = ast[adr]; (s)% ⊦ ¬ n ∈ st ∧ adr ∈ st → adr ∈ st[n, ablock]; (ss)% ⊦ adr ≠ adr0 → (adr ∈ ast[adr0, a0] ↔ adr ∈ ast); (s)