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Proof step 12 from proof at-store-inv (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
st[n, mkblock(x, lax, m)][n0] = mkblock(x0, lax0, m0), inv-flashsize(x, lax), n0 ∈ st[n, mkblock(x, lax, m)], ∀ n0. n0 ∈ st → # st[n0].list = # st[n0].ladrlist ∧ inv-flashsize(st[n0].list, st[n0].ladrlist) ∧ ¬ dups(st[n0].ladrlist) ⊦ dups(lax), inv-flashsize(x0, lax0) ∧ ¬ dups(lax0)

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(x, lax, m)][n0], x0, lax0, m0]

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(x, lax, m)][n0], x0, lax0, m0