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Proof step 16 from proof remove-refsize0-subset-refed (previous step)

st[m] = mkblock(x, lax, m2), @(m, m0, m1) ∈ st, m ∈ st, @(m, m0, m1) ∈ nodes(st), @(m, m0, m1) ∈ as, as ⊆ dom nodes(st), ∀ n. n ∈ st → st[n].ref-size = (as | n).size, ∀ adr. adr ∈ as → adr.size ≠ 0 ⊦ # as = 0, nodes(st)[@(m, m0, m1)] = remove-refsize0(st)[@(m, m0, m1)], as = ∅, m2 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
st[m] = mkblock(x, lax, m2), m ∈ st, @(m, m0, m1) ∈ as, @(m, m0, m1) ∈ st, as ⊆ dom nodes(st), ∀ adr. adr ∈ as → adr.size ≠ 0, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, st[@(m, m0, m1)] = remove-refsize0(st)[@(m, m0, m1)], as = ∅, m2 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ nodes(st) ↔ adr ∈ st; (s)% ⊦ adr ∈ st → nodes(st)[adr] = st[adr]; (s)