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Proof step 22 from proof remove-refsize0-subset-refed (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 23:
st[m] = mkblock(x, lax, m2), remove-refsize0(st)[m] = mkblock(x0, lax0, m3), @(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)] = x0[pos(@(m0, m1), lax0)], as = ∅, m2 = 0

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] → [remove-refsize0(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: remove-refsize0(st)[m], x0, lax0, m3