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Proof step 11 from proof in-remove-refsize0-refsize-cons (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
(∅)[m] = mkblock(x, lax, m2), st[m] = mkblock(x0, lax0, m3), m3 = 0, inv(st), m ∈ st, @(m, m0, m1) ∈ as, @(m, m0, m1) ∈ st, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅

Rule argument

spec: ablock name: elim seq: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m) subst: [ablock, x, lax, m] → [st[m], x0, lax0, m3]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st[m], x0, lax0, m3