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Proof step 21 from proof remove-refsize0-subset-refed (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
st[m] = mkblock(x, lax, m2), @(m, m0, m1) ∈ remove-refsize0(st), 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].list[pos(@(m0, m1), remove-refsize0(st)[m].ladrlist)], as = ∅, m2 = 0

Rule argument

Additional information

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