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Proof step 8 from proof refsize-cons-write (previous step)

st[n] = mkblock(x0, lax0, m0), ¬ n ∈ st ∨ m0 = (as | n).size → m = (as | n).size, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ ∀ n0. n0 ∈ st[n, mkblock(x, lax, m)] → st[n, mkblock(x, lax, m)][n0].ref-size = (as | n0).size

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
st[n] = mkblock(x0, lax0, m0), n0 ∈ st[n, mkblock(x, lax, m)], ¬ n ∈ st ∨ m0 = (as | n).size → m = (as | n).size, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ st[n, mkblock(x, lax, m)][n0].ref-size = (as | n0).size

Proof step 8 from proof refsize-cons-write (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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