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Proof step 6 from proof prefix-alloc (previous step)

n1 ∈ st, n1 ∈ st[n, mkblock(x, lax, n0)], ∀ n1. n1 ∈ st → # st[n1].list = # st[n1].ladrlist ∧ inv-flashsize(st[n1].list, st[n1].ladrlist) ∧ ¬ dups(st[n1].ladrlist) ⊦ # st[n1].list = # st[n1].ladrlist, n ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
st[n1] = mkblock(x0, lax0, m), n1 ∈ st[n, mkblock(x, lax, n0)], n1 ∈ st, ∀ n1. n1 ∈ st → # st[n1].list = # st[n1].ladrlist ∧ inv-flashsize(st[n1].list, st[n1].ladrlist) ∧ ¬ dups(st[n1].ladrlist) ⊦ # x0 = # lax0, 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[n1], x0, lax0, 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[n1], x0, lax0, m