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Proof step 14 from proof nodes-alloc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
st[m0] = mkblock(x, lax, m3), m0 ∈ st, @(m1, m2) ∈ lax ⊦ st[n, mkblock([], [], m)][@(m0, m1, m2)] = nodes(st)[@(m0, m1, m2)], n ∈ st

Rule argument

Additional information

Used simplifier rules: ⊦ st[n] = mkblock(x, lax, m0) ∧ n ∈ st ∧ @(n0, n1) ∈ lax → @(n, n0, n1) ∈ st; (ss)