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Proof step 4 from proof refsize-cons-write (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ 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) ↔ ¬ n ∈ st ∨ st[n].ref-size = (as | n).size → m = (as | n).size

Proof step 4 from proof refsize-cons-write (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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