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Proof step 35 from proof remove-refsize0-subset-refed (previous step)

(∅)[m] = mkblock(x, lax, m2), st[m] = mkblock(x0, lax0, 0), m ∈ st, @(m, m0, m1) ∈ st, @(m, m0, m1) ∈ as, as ⊆ dom nodes(st), @(m, m0, m1) ∈ as → @(m, m0, m1).size ≠ 0, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
st[m] = mkblock(x0, lax0, 0), (∅)[m] = mkblock(x, lax, m2), @(m, m0, m1) ∈ as, @(m, m0, m1) ∈ st, m ∈ st, as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, m1 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ @(m, m0, m1).size = m1; (s)