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Proof step 3 from proof refsize-cons-remove-index (previous step)

st[m] = mkblock(x, lax, n0 + m1), n ∈ st[m, mkblock(x, lax, n0)], @(m, m0, m1) ∈ as, m ∈ st, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, st[m, mkblock(x, lax, n0)][n].ref-size = ((as -- @(m, m0, m1)) | n).size, as = ∅

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
st[m] = mkblock(x, lax, n0 + m1), st[m, mkblock(x, lax, n0)][n] = mkblock(x0, lax0, m2), m ∈ st, @(m, m0, m1) ∈ as, n ∈ st[m, mkblock(x, lax, n0)], ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, m2 = ((as -- @(m, m0, m1)) | n).size

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[m, mkblock(x, lax, n0)][n], x0, lax0, m2]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st[m, mkblock(x, lax, n0)][n], x0, lax0, m2