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Proof step 3 from proof refsize-cons-alloc (previous step)

as | n = ∅, n0 ∈ st[n, mkblock(a ', @(m0, m1) ', m1)], ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ st[n, mkblock(a ', @(m0, m1) ', m1)][n0].ref-size = ((as ++ @(n, m0, m1)) | n0).size, n ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
st[n, mkblock(a ', @(m0, m1) ', m1)][n0] = mkblock(x, lax, m), as | n = ∅, n0 ∈ st[n, mkblock(a ', @(m0, m1) ', m1)], ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m = ((as ++ @(n, m0, m1)) | n0).size, n ∈ st

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[n, mkblock(a ', @(m0, m1) ', m1)][n0], x, lax, m]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st[n, mkblock(a ', @(m0, m1) ', m1)][n0], x, lax, m