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Proof step 28 from proof remove-refsize0-subset-refed (previous step)

remove-refsize0(st)[m] = mkblock(x, lax, m2), @(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, as = ∅, m ∈ remove-refsize0(st), m ∈ remove-refsize0(st) ∧ @(m0, m1) ∈ lax ∧ (nodes(st) | as)[@(m, m0, m1)] = remove-refsize0(st)[@(m, m0, m1)]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
(∅)[m] = mkblock(x, lax, m2), @(m, m0, m1) ∈ 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, as = ∅, m ∈ remove-refsize0(st)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n ∈ st → st[n] = (∅)[n]; (s)