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Proof step 19 from proof remove-refsize0-subset-refed (previous step)

st[m] = mkblock(x, lax, m2), @(m, m0, m1) ∈ st, @(m, m0, m1) ∈ as, m ∈ 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)], as = ∅, m2 = 0, @(m, m0, m1) ∈ remove-refsize0(st)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: ablock-store-inv name: in seq: ⊦ adr ∈ st ↔ adr.lnum ∈ st ∧ @(adr.pos, adr.size) ∈ st[adr.lnum].ladrlist subst: [st, adr] → [remove-refsize0(st), @(m, m0, m1)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ @(m, m0, m1).lnum = m; (s)% ⊦ n ∈ remove-refsize0(st) ↔ n ∈ st ∧ st[n].ref-size ≠ 0; (ws)% ⊦ @(m, m0, m1).pos = m0; (s)% ⊦ @(m, m0, m1).size = m1; (s)% ⊦ n ∈ st ∧ st[n].ref-size ≠ 0 → remove-refsize0(st)[n] = st[n]; (s)% ⊦ mkblock(x, lax, m).ref-size = m; (s)% ⊦ mkblock(x, lax, m).ladrlist = lax; (s)% ⊦ st[n] = mkblock(x, lax, m0) ∧ @(n, n0, n1) ∈ st → @(n0, n1) ∈ lax; (ss) Used terms: remove-refsize0(st), @(m, m0, m1)