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Proof step 8 from proof in-remove-refsize0-refsize-cons (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ n ∈ st ∧ st[n].ref-size ≠ 0 → remove-refsize0(st)[n] = st[n]
Proof step 10:
remove-refsize0(st)[m] = mkblock(x, lax, m2), inv(st), @(m, m0, m1) ∈ st, @(m, m0, m1) ∈ as, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, m ∈ remove-refsize0(st) ∧ @(m0, m1) ∈ lax, m ∈ st ∧ st[m].ref-size ≠ 0

Rule argument

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

Additional information

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