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Proof step 7 from proof in-remove-refsize0-refsize-cons (previous step)

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) ∈ remove-refsize0(st)[m].ladrlist

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
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

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], x, lax, m2]

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], x, lax, m2