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Proof step 29 from proof remove-refsize0-subset-refed (previous step)

(∅)[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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ n ∈ remove-refsize0(st) ↔ n ∈ st ∧ st[n].ref-size ≠ 0
Proof step 31:
(∅)[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 ∈ st ∧ st[m].ref-size ≠ 0

Rule argument

spec: ablock-store-inv name: in-remove-refsize0 seq: ⊦ n ∈ remove-refsize0(st) ↔ n ∈ st ∧ st[n].ref-size ≠ 0 subst: [st, n] → [st, m] flags: <all paths>

Additional information

Used terms: st, m